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Read More## MCQs on electrostatic potential & capacitance

**Question 1: The electrostatic potential at a point in an electric field is 200 V. What does this indicate? **

A) The magnitude of the electric field at that point B) The work done to bring a unit positive charge from infinity to that point C) The electric potential energy at that point D) The magnitude of the electric field due to a nearby charge

Answer: B) The work done to bring a unit positive charge from infinity to that point

**Question 2: Two capacitors, C1 and C2, are connected in parallel. If C1 has a capacitance of 4 μF and C2 has a capacitance of 6 μF, what is the total equivalent capacitance? **

A) 2 μF B) 10 μF C) 24 μF D) 8 μF

Answer: B) 10 μF

**Question 3: The electric potential energy stored in a capacitor is given by: **

A) U = CV B) U = 1/2 CV^2 C) U = Q^2/2C D) U = 1/2 Q^2C

Answer: D) U = 1/2 Q^2C

**Question 4: A capacitor is connected to a battery. If the battery is disconnected and the distance between the plates of the capacitor is doubled, what happens to the stored energy? **

A) It doubles B) It halves C) It quadruples D) It remains the same

Answer: B) It halves

**Question 5: The capacitance of a parallel-plate capacitor depends on:**

A) The separation between the plates B) The area of the plates C) The dielectric constant of the material between the plates D) All of the above

Answer: D) All of the above

**6. The potential energy of a pair of charges is given by U = kqq/r, where k is a constant, q is the charge of each particle, and r is the distance between them. If the charge of one particle is doubled, what happens to the potential energy?**

(A) It is doubled. (B) It is halved. (C) It is quadrupled. (D) It is unchanged.

**Answer: (A)**

The potential energy is proportional to the square of the charge. Doubling the charge will quadruple the potential energy.

**7. Two charges of +10μC and -10μC are placed 10cm apart. What is the electric field at a point halfway between them?**

(A) 100N/C (B) 500N/C (C) -500N/C (D) 0N/C

**Answer: (C)**

The electric field is directed from the positive charge to the negative charge. The magnitude of the electric field is given by E = kqq/r^2, where k is a constant, q is the charge of each particle, and r is the distance between them. In this case, r = 10cm/2 = 5cm. The electric field is therefore E = 9×10^9 N⋅m^2/C^2 * 10μC * (-10μC) / (5cm)^2 = -500N/C.

**8. A parallel plate capacitor has plates of area 100cm^2 and a separation of 1mm. The capacitance of the capacitor is 10μF. What is the dielectric constant of the material between the plates?**

(A) 1 (B) 10 (C) 100 (D) 1000

**Answer: (C)**

The capacitance of a parallel plate capacitor is given by C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of the plates, and d is the separation between the plates. The dielectric constant of a material is a measure of how much it increases the capacitance of a capacitor. In this case, the dielectric constant is ε = C/ε₀A/d = 10μF / (8.85 x 10^-12 F/m * 100cm^2 / 1mm) = 100.

**9. A capacitor is charged to a potential difference of 10V. If the charge on the capacitor is 10μC, what is the capacitance of the capacitor?**

(A) 1μF (B) 10μF (C) 100μF (D) 1000μF

**Answer: (B)**

The charge on a capacitor is given by Q = CV, where C is the capacitance of the capacitor and V is the potential difference across the capacitor. In this case, Q = 10μC and V = 10V. The capacitance is therefore C = Q/V = 10μC / 10V = 1μF.

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Read MoreMCQs on electrostatic potential & capacitance MCQs on electrostatic potential & capacitance MCQs on electrostatic potential & capacitance MCQs on electrostatic potential & capacitance MCQs on electrostatic potential & capacitance MCQs on electrostatic potential & capacitance MCQs on electrostatic potential & capacitance MCQs on electrostatic potential & capacitance MCQs on electrostatic potential & capacitance MCQs on electrostatic potential & capacitance MCQs on electrostatic potential & capacitance