A point charge q is released from rest at point A and accelerates in a uniform e

Question

A point charge q is released from rest at point A and accelerates in a uniform electric field E. What is the ratio between the work done by the field on the charge: W_A rightarrow B/W_B rightarrow C (A.) 1/2(B) 1/4 (C) 1 (D) 2/1 (E) 4/1 A parallel-plate capacitor is charged by connection to a battery and remains connected. What will happen to the charge on the capacitor and voltage across it if the separation between the plates is decreased and the area is increased? (A) Both increase (B) Both decrease (C) Both remain the same (D) The voltage remains the same and charge increases (E) The voltage remains the same and charge decreases

Explanation / Answer

( 3 ). The amount of work done to move charge is called potential. As both B and C are equvipotetial surfaces and equally separted , the work done by the field is same as they move from A to B and work done by the field as move from B to C.

Thus, the ratio of work done is 1

Correct option: ( C )

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( 2 )

The capacitance of parallel plate capacitor is

C = A o / d

As separation decreased and area increased, the capacitance increased.

Charge on capacitor:

Q = C V

As source connected ( battery ) is same, the voltage same and hence the charge on capacitor increses as capacitance increased.

Correct option is ( D )

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