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2 years ago

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Find the voltage change when: a. An electric field does 12 J of work on a 0.0001-C charge. b. The same electric field does 24 J

of work on a 0.0002-C charge.

1 answer:

kondaur [170]2 years ago

7 0

Explanation:

Given that,

(a) Work done by the electric field is 12 J on a 0.0001 C of charge. The electric potential is defined as the work done per unit charged particles. It is given by :

$V=\dfrac{W}{q}$

$V=\dfrac{12}{0.0001}$

$V=12\times 10^4\ Volt$

(b) Similarly, same electric field does 24 J of work on a 0.0002-C charge. The electric potential difference is given by :

$V=\dfrac{W}{q}$

$V=\dfrac{24}{0.0002}$

$V=12\times 10^4\ Volt$

Therefore, this is the required solution.

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and

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