Question
A parallel–plate capacitor has a capacitance Co. A second parallel–plate capacitor has plates with twice the area and twice the separation. The capacitance of the second capacitor is most nearly
(A) ¼Co (B) Co (C) 2Co (D) 4Co
▶️Answer/Explanation
Ans:B
Solution: C = ε0A/d; if A × 2, C × 2 and if d × 2, C ÷ 2 so the net effect is C is unchanged
Question
The capacitance of a parallel–plate capacitor can be increased by increasing which of the following?
(A) The distance between the plates (B) The charge on each plate (C) The area of the plates
(D) The potential difference across the plates
▶️Answer/Explanation
Ans:C
Solution: C = ε0A/d and changing Q or V has no effect on the capacitance
Question
A parallel–plate capacitor has a capacitance Co. A second parallel–plate capacitor has plates with twice the area and twice the separation. The capacitance of the second capacitor is most nearly
(A) ¼Co (B) ½Co (C) Co (D) 2Co
▶️Answer/Explanation
Ans:C
Solution: C = ε0A/d; ε0(2A)/(2d) = ε0A/d
Question
Which of the following statements about conductors under electrostatic conditions is true?
(A) Positive work is required to move a positive charge over the surface of a conductor.
(B) Charge that is placed on the surface of a conductor always spreads evenly over the surface.
(C) The electric potential inside a conductor is always zero.
(D) The surface of a conductor is always an equipotential surface.
▶️Answer/Explanation
Ans:D
Solution: Since there is no component of the electric field along a conducting surface under electrostatic conditions, no work is done moving the charge around the surface, meaning no differences in potential
Question
A sheet of mica is inserted between the plates of an isolated charged parallel–plate capacitor. Which of the following statements is true?
(A) The capacitance decreases.
(B) The potential difference across the capacitor decreases.
(C) The charge on the capacitor plates decreases
(D) The electric field between the capacitor plates increases.
▶️Answer/Explanation
Ans:B
Solution: Since the capacitor is isolated, Q remains constant. Filling the place with oil (a dielectric) will increase the capacitance, causing the potential (V = Q/C) to decrease.