Electrostatics is a fundamental chapter in Physics that focuses on the study of stationary electric charges and their interactions. This chapter introduces students to the basic principles of electric forces, fields, and potentials, providing a comprehensive understanding of how electric charges interact when at rest. The unit covers the concepts of Coulomb’s Law, which describes the force between two point charges, and explores the electric field and electric potential associated with these charges. It also delves into the properties of conductors and insulators, and the principles of electric shielding.
This chapter is vital for understanding the principles of electric charges and fields, which are foundational to many areas of Physics and practical technologies. Mastering Electrostatics is essential for academic success and for applying these concepts in real-world scenarios.
a) Ampere (A)
b) Volt (V)
c) Coulomb (C)
d) Ohm (Ω)
Answer: c) Coulomb (C)
a) Coulomb’s Law
b) Ohm’s Law
c) Newton’s Law
d) Faraday’s Law
Answer: a) Coulomb’s Law
a) Inversely proportional to the square of the distance between them
b) Directly proportional to the product of their charges
c) Independent of the distance between them
d) Both a) and b)
Answer: d) Both a) and b)
a) E=kQr2E = \frac{kQ}{r^2}E=r2kQ
b) E=Q4πϵ0r2E = \frac{Q}{4\pi \epsilon_0 r^2}E=4πϵ0r2Q
c) E=FQE = \frac{F}{Q}E=QF
d) E=kQE = \frac{k}{Q}E=Qk
Answer: b) E=Q4πϵ0r2E = \frac{Q}{4\pi \epsilon_0 r^2}E=4πϵ0r2Q
a) Newton per Coulomb (N/C)
b) Coulomb per Newton (C/N)
c) Joule per Coulomb (J/C)
d) Volt per meter (V/m)
Answer: a) Newton per Coulomb (N/C)
a) Work done per unit charge in bringing a positive test charge from infinity to that point
b) Force experienced by a unit charge at that point
c) Energy stored per unit volume at that point
d) Charge per unit area at that point
Answer: a) Work done per unit charge in bringing a positive test charge from infinity to that point
a) Volt (V)
b) Ampere (A)
c) Coulomb (C)
d) Ohm (Ω)
Answer: a) Volt (V)
a) V=kQrV = \frac{kQ}{r}V=rkQ
b) V=Q4πϵ0rV = \frac{Q}{4\pi \epsilon_0 r}V=4πϵ0rQ
c) V=Qr2V = \frac{Q}{r^2}V=r2Q
d) V=krV = \frac{k}{r}V=rk
Answer: b) V=Q4πϵ0rV = \frac{Q}{4\pi \epsilon_0 r}V=4πϵ0rQ
a) E=−dVdrE = -\frac{dV}{dr}E=−drdV
b) E=dVdrE = \frac{dV}{dr}E=drdV
c) E=VrE = \frac{V}{r}E=rV
d) E=−VrE = -\frac{V}{r}E=−rV
Answer: a) E=−dVdrE = -\frac{dV}{dr}E=−drdV
a) W=QVW = QVW=QV
b) W=QVrW = \frac{QV}{r}W=rQV
c) W=VQW = \frac{V}{Q}W=QV
d) W=QVW = \frac{Q}{V}W=VQ
Answer: a) W=QVW = QVW=QV
a) The charge stored per unit potential difference
b) The potential difference per unit charge
c) The work done per unit charge
d) The energy stored per unit charge
Answer: a) The charge stored per unit potential difference
a) Farad (F)
b) Volt (V)
c) Ampere (A)
d) Coulomb (C)
Answer: a) Farad (F)
a) C=ϵ0AdC = \frac{\epsilon_0 A}{d}C=dϵ0A
b) C=dϵ0AC = \frac{d}{\epsilon_0 A}C=ϵ0Ad
c) C=Aϵ0dC = \frac{A}{\epsilon_0 d}C=ϵ0dA
d) C=ϵ0dAC = \frac{\epsilon_0 d}{A}C=Aϵ0d
Answer: a) C=ϵ0AdC = \frac{\epsilon_0 A}{d}C=dϵ0A
a) E=12CV2E = \frac{1}{2} CV^2E=21CV2
b) E=12QVE = \frac{1}{2} QVE=21QV
c) E=12Q2CE = \frac{1}{2} \frac{Q^2}{C}E=21CQ2
d) All of the above
Answer: d) All of the above
a) 1Ceq=1C1+1C2+⋯+1Cn\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n}Ceq1=C11+C21+⋯+Cn1
b) Ceq=C1+C2+⋯+CnC_{eq} = C_1 + C_2 + \cdots + C_nCeq=C1+C2+⋯+Cn
c) Ceq=C1C2C1+C2C_{eq} = \frac{C_1 C_2}{C_1 + C_2}Ceq=C1+C2C1C2
d) Ceq=C1+C2C1C2C_{eq} = \frac{C_1 + C_2}{C_1 C_2}Ceq=C1C2C1+C2
Answer: a) 1Ceq=1C1+1C2+⋯+1Cn\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n}Ceq1=C11+C21+⋯+Cn1
a) Ceq=C1+C2+⋯+CnC_{eq} = C_1 + C_2 + \cdots + C_nCeq=C1+C2+⋯+Cn
b) 1Ceq=1C1+1C2+⋯+1Cn\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n}Ceq1=C11+C21+⋯+Cn1
c) Ceq=C1C2C1+C2C_{eq} = \frac{C_1 C_2}{C_1 + C_2}Ceq=C1+C2C1C2
d) Ceq=C1+C2C1C2C_{eq} = \frac{C_1 + C_2}{C_1 C_2}Ceq=C1C2C1+C2
Answer: a) Ceq=C1+C2+⋯+CnC_{eq} = C_1 + C_2 + \cdots + C_nCeq=C1+C2+⋯+Cn
a) U=14πϵ0∑qiqjrijU = \frac{1}{4\pi \epsilon_0} \sum \frac{q_i q_j}{r_{ij}}U=4πϵ01∑rijqiqj
b) U=12QVU = \frac{1}{2} QVU=21QV
c) U=12CV2U = \frac{1}{2} CV^2U=21CV2
d) U=14πϵ0∑qiqjrij2U = \frac{1}{4\pi \epsilon_0} \sum \frac{q_i q_j}{r_{ij}^2}U=4πϵ01∑rij2qiqj
Answer: a) U=14πϵ0∑qiqjrijU = \frac{1}{4\pi \epsilon_0} \sum \frac{q_i q_j}{r_{ij}}U=4πϵ01∑rijqiqj
a) Φ=Qencϵ0\Phi = \frac{Q_{enc}}{\epsilon_0}Φ=ϵ0Qenc
b) Φ=E⋅A\Phi = E \cdot AΦ=E⋅A
c) Φ=E⋅Aϵ0\Phi = \frac{E \cdot A}{\epsilon_0}Φ=ϵ0E⋅A
d) Φ=Qenc⋅Aϵ0\Phi = \frac{Q_{enc} \cdot A}{\epsilon_0}Φ=ϵ0Qenc⋅A
Answer: a) Φ=Qencϵ0\Phi = \frac{Q_{enc}}{\epsilon_0}Φ=ϵ0Qenc
a) The electric field intensity at the surface
b) The surface area of the closed surface
c) The charge enclosed within the surface
d) The potential difference across the surface
Answer: c) The charge enclosed within the surface
a) Zero
b) Maximum at the surface
c) Constant throughout
d) Increasing towards the center
Answer: a) Zero
a) The work done in moving a unit positive charge from one point to another
b) The force experienced by a unit positive charge
c) The product of the charge and the distance
d) The work done in moving a unit charge from infinity to one point
Answer: a) The work done in moving a unit positive charge from one point to another
a) σ2ϵ0\frac{\sigma}{2\epsilon_0}2ϵ0σ
b) σ4πϵ0\frac{\sigma}{4\pi \epsilon_0}4πϵ0σ
c) σϵ0\frac{\sigma}{\epsilon_0}ϵ0σ
d) Q4πϵ0r2\frac{Q}{4\pi \epsilon_0 r^2}4πϵ0r2Q
Answer: c) σϵ0\frac{\sigma}{\epsilon_0}ϵ0σ
a) λ2πϵ0lnr\frac{\lambda}{2 \pi \epsilon_0} \ln r2πϵ0λlnr
b) λ4πϵ0r\frac{\lambda}{4 \pi \epsilon_0 r}4πϵ0rλ
c) λ4πϵ0r2\frac{\lambda}{4 \pi \epsilon_0 r^2}4πϵ0r2λ
d) λr4πϵ0\frac{\lambda r}{4 \pi \epsilon_0}4πϵ0λr
Answer: a) λ2πϵ0lnr\frac{\lambda}{2 \pi \epsilon_0} \ln r2πϵ0λlnr
a) Parallel to the surface
b) Perpendicular to the surface
c) Zero everywhere
d) Always directed towards the center
Answer: b) Perpendicular to the surface
a) C=4πϵ0r1r2r2−r1C = 4\pi \epsilon_0 \frac{r_1 r_2}{r_2 – r_1}C=4πϵ0r2−r1r1r2
b) C=4πϵ0r2−r1r1r2C = 4\pi \epsilon_0 \frac{r_2 – r_1}{r_1 r_2}C=4πϵ0r1r2r2−r1
c) C=4πϵ0r1r2r1+r2C = 4\pi \epsilon_0 \frac{r_1 r_2}{r_1 + r_2}C=4πϵ0r1+r2r1r2
d) C=4πϵ0(r2−r1)C = 4\pi \epsilon_0 (r_2 – r_1)C=4πϵ0(r2−r1)
Answer: a) C=4πϵ0r1r2r2−r1C = 4\pi \epsilon_0 \frac{r_1 r_2}{r_2 – r_1}C=4πϵ0r2−r1r1r2
a) U=12CV2U = \frac{1}{2} CV^2U=21CV2
b) U=QVU = QVU=QV
c) U=12Q2/CU = \frac{1}{2} Q^2/CU=21Q2/C
d) All of the above
Answer: d) All of the above
a) Zero
b) Equal to the field outside the sphere
c) Maximum at the center
d) Varies inversely with the distance from the center
Answer: a) Zero
a) Q=CVQ = CVQ=CV
b) V=QCV = \frac{Q}{C}V=CQ
c) C=QVC = \frac{Q}{V}C=VQ
d) All of the above
Answer: d) All of the above
a) Always attractive
b) Always repulsive
c) Depends on the medium between them
d) Always zero
Answer: c) Depends on the medium between them
a) Uniform
b) Radially outward or inward
c) Constant in magnitude
d) Zero at all points
Answer: b) Radially outward or inward
a) Always start from positive charges and end on negative charges
b) Always form closed loops
c) Are always parallel to each other
d) Can cross each other
Answer: a) Always start from positive charges and end on negative charges
a) The product of the charges
b) The distance between the charges
c) The square of the distance between the charges
d) The cube of the distance between the charges
Answer: c) The square of the distance between the charges
a) The ratio of the electric field without dielectric to the electric field with dielectric
b) The ratio of the capacitance with dielectric to the capacitance without dielectric
c) The ratio of the charge with dielectric to the charge without dielectric
d) The ratio of the energy stored with dielectric to the energy stored without dielectric
Answer: b) The ratio of the capacitance with dielectric to the capacitance without dielectric
a) The work done in moving a positive charge from infinity to that point
b) The potential energy per unit charge at that point
c) The force per unit charge at that point
d) The work done per unit charge in moving a charge between two points
Answer: b) The potential energy per unit charge at that point
a) The square of the distance from the dipole
b) The cube of the distance from the dipole
c) The fourth power of the distance from the dipole
d) The distance from the dipole
Answer: b) The cube of the distance from the dipole
a) The electric potential at that point
b) The potential energy at that point
c) The electric field intensity at that point
d) The electric flux through a surface enclosing that point
Answer: b) The potential energy at that point
a) Non-zero and constant
b) Zero in electrostatic equilibrium
c) Maximum at the center
d) Dependent on the external electric field
Answer: b) Zero in electrostatic equilibrium
a) F=qEF = qEF=qE
b) F=EqF = \frac{E}{q}F=qE
c) F=E⋅qϵ0F = \frac{E \cdot q}{\epsilon_0}F=ϵ0E⋅q
d) F=qEF = \frac{q}{E}F=Eq
Answer: a) F=qEF = qEF=qE
a) The work done per unit charge in moving a charge from one point to the other
b) The force per unit charge at one of the points
c) The energy stored in the electric field between the two points
d) The sum of the electric field intensities at the two points
Answer: a) The work done per unit charge in moving a charge from one point to the other
a) E=Q4πϵ0r2E = \frac{Q}{4\pi \epsilon_0 r^2}E=4πϵ0r2Q
b) E=Q4πϵ0R2E = \frac{Q}{4\pi \epsilon_0 R^2}E=4πϵ0R2Q
c) E=Q4πϵ0rR3E = \frac{Q}{4\pi \epsilon_0} \frac{r}{R^3}E=4πϵ0QR3r
d) E=Q4πϵ0r2R3E = \frac{Q}{4\pi \epsilon_0} \frac{r^2}{R^3}E=4πϵ0QR3r2
Answer: c) E=Q4πϵ0rR3E = \frac{Q}{4\pi \epsilon_0} \frac{r}{R^3}E=4πϵ0QR3r
a) Same as if all the charge were concentrated at the center
b) Zero
c) Directly proportional to the distance from the center
d) Inversely proportional to the distance from the center
Answer: a) Same as if all the charge were concentrated at the center
a) The area of the plates
b) The separation between the plates
c) The dielectric material between the plates
d) All of the above
Answer: d) All of the above
a) The points are at the same electric potential
b) The points are at different electric potentials
c) The electric field between the points is maximum
d) The electric field between the points is minimum
Answer: a) The points are at the same electric potential
a) V=kQr2+z2V = \frac{kQ}{\sqrt{r^2 + z^2}}V=r2+z2kQ
b) V=kQr2+z2V = \frac{kQ}{r^2 + z^2}V=r2+z2kQ
c) V=kQr2−z2V = \frac{kQ}{\sqrt{r^2 – z^2}}V=r2−z2kQ
d) V=kQr2−z2V = \frac{kQ}{r^2 – z^2}V=r2−z2kQ
Answer: a) V=kQr2+z2V = \frac{kQ}{\sqrt{r^2 + z^2}}V=r2+z2kQ
a) It helps in calculating the electric field
b) It simplifies the calculation of work done in moving charges
c) It provides a measure of the energy stored in an electric field
d) All of the above
Answer: d) All of the above
a) The distance between the charges is doubled
b) The distance between the charges is halved
c) The charges are doubled
d) The charges are halved
Answer: a) The distance between the charges is doubled
a) Point radially inward
b) Point radially outward
c) Are circular around the charge
d) Are parallel lines
Answer: b) Point radially outward
a) Zero
b) Same as outside the shell
c) Constant and equal to the potential at the surface
d) Varies with the distance from the center
Answer: c) Constant and equal to the potential at the surface
a) Electric field lines
b) Electric potential
c) Capacitance
d) Dielectric constant
Answer: a) Electric field lines
a) The initial and final positions of the charge
b) The path taken by the charge
c) The medium through which the charge is moved
d) The type of charge being moved
Answer: a) The initial and final positions of the charge
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