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Rajasthan Board Class 12 (RBSE) Electrostatic Potential and Capacitance — practice questions

Explore 23 free Electrostatic Potential and Capacitance questions in Rajasthan Board Class 12 (RBSE) Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Electrostatic Potential and Capacitance questions and exercises with answers

Showing questions 1–23 of 23.

Electric potential is a:Read the answer and worked solution →An equipotential surface is perpendicular to:Read the answer and worked solution →Capacitance of a parallel-plate capacitor is:Read the answer and worked solution →Energy stored in a capacitor equals:Read the answer and worked solution →Capacitors in series share:Read the answer and worked solution →Time constant of an RC circuit is:Read the answer and worked solution →The SI unit of electric potential isRead the answer and worked solution →Electric potential at infinity in the standard syllabus convention is taken asRead the answer and worked solution →The SI unit of capacitance isRead the answer and worked solution →Electric field at any point inside a hollow charged conductor isRead the answer and worked solution →On an equipotential surface, the work done in moving a test charge between any two points isRead the answer and worked solution →The potential at a distance 0.30 m from a point charge of +20 nC is (k = 9 × 10⁹ N m² C⁻²)Read the answer and worked solution →A parallel-plate capacitor in vacuum has plate area A and separation d. Its capacitance isRead the answer and worked solution →Three capacitors of 2 µF each are connected in series. The equivalent capacitance isRead the answer and worked solution →The work done in bringing a charge q from infinity to a point where the potential is V isRead the answer and worked solution →Two capacitors of 4 µF and 6 µF are connected in parallel and charged by a 10 V battery. The total charge stored isRead the answer and worked solution →When the separation between the plates of an air parallel-plate capacitor is doubled (keeping plate area constant), the capacitanceRead the answer and worked solution →An electric dipole of moment p is placed in a uniform field E. The work done to rotate it from the orientation θ = 0 (aligned with the field) to θ = 180° isRead the answer and worked solution →A 5 µF capacitor is charged to 200 V. The energy stored in it isRead the answer and worked solution →An isolated conducting sphere of radius R carries a charge Q. The electric potential at its centre is (V(∞) = 0)Read the answer and worked solution →Four equal point charges q sit at the corners of a square of side d. The work needed to assemble this configuration from infinity isRead the answer and worked solution →Match each electrostatic termRead the answer and worked solution →True or false: electrostaticsRead the answer and worked solution →

Worked practice questions

Electric potential is a:

  1. Scalar with unit volt
  2. Vector
  3. Field line
  4. Force
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Correct answer: A. Scalar with unit volt

### Step-by-Step Solution **1. Definition of Electric Potential:** Electric potential ($V$) at a point in an electric field is defined as the amount of work done ($W$) per unit positive test charge ($q$) in bringing the charge from infinity to that point without acceleration: $$V = \frac{W}{q}$$ **2. Determining the Type of Quantity:** - **Work ($W$):** Work is a scalar quantity because it is the dot product of force and displacement. - **Charge ($q$):** Charge is a fundamental scalar quantity. Since the ratio of two scalars is itself a scalar, **Electric Potential is a scalar quantity**. It has magnitude but no direction. **3. Units of Electric Potential:** The SI unit for work is the Joule ($J$) and for charge is the Coulomb ($C$). Therefore, the unit for potential is: $$\text{Unit of } V = \frac{\text{Joule}}{\text{Coulomb}} = \text{J/C}$$ In the International System of Units (SI), $1 \text{ J/C}$ is defined as **$1 \text{ Volt (V)}$**. **Conclusion:** Electric potential is a scalar quantity with the unit Volt.

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An equipotential surface is perpendicular to:

  1. The flow of charge
  2. The electric field lines
  3. Other equipotentials
  4. The time axis
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Correct answer: B. The electric field lines

An equipotential surface is defined as a surface where every point has the same electric potential ($V = \text{constant}$). 1. **Work and Potential Difference**: The work done ($W$) in moving a charge $q$ from one point to another on a surface is given by $W = q(V_2 - V_1)$. 2. **Zero Work**: Since $V_1 = V_2$ for any two points on an equipotential surface, the potential difference $\Delta V = 0$. Therefore, the work done in moving a charge along an equipotential surface is zero ($W = 0$). 3. **Force and Displacement**: Work is also defined by the dot product of the electric force ($\vec{F} = q\vec{E}$) and the displacement vector ($d\vec{s}$): $$W = \int q\vec{E} \cdot d\vec{s} = 0$$ 4. **Perpendicularity**: For the dot product $\vec{E} \cdot d\vec{s}$ to be zero for any displacement $d\vec{s}$ along the surface (where $\vec{E} \neq 0$), the electric field vector $\vec{E}$ must be perpendicular to the displacement vector. Conclusion: Since electric field lines represent the direction of the electric field vector at any point, the electric field lines must be perpendicular to the equipotential surface.

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Capacitance of a parallel-plate capacitor is:

  1. ε₀d/A
  2. ε₀A·d
  3. ε₀A/d
  4. A/(ε₀d)
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Correct answer: C. ε₀A/d

To derive the capacitance of a parallel-plate capacitor, we follow these steps: 1. **Define Capacitance**: Capacitance $C$ is the ratio of the magnitude of the charge $Q$ on either conductor to the magnitude of the potential difference $V$ between them: $$C = \frac{Q}{V}$$ 2. **Find the Electric Field**: For a parallel-plate capacitor with area $A$ and charge $Q$, the surface charge density is $\sigma = Q/A$. The electric field $E$ between the plates is uniform and given by: $$E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}$$ 3. **Relate Field to Potential**: The potential difference $V$ between plates separated by distance $d$ is: $$V = E \cdot d = \frac{Q \cdot d}{\epsilon_0 A}$$ 4. **Calculate Capacitance**: Substitute $V$ back into the capacitance formula: $$C = \frac{Q}{V} = \frac{Q}{\left(\frac{Q \cdot d}{\epsilon_0 A}\right)} = \frac{\epsilon_0 A}{d}$$ Thus, the capacitance of a parallel-plate capacitor is $C = \frac{\epsilon_0 A}{d}$.

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