Question

• How much work will be done in bringing a charge
of 5 millicoulombs from infinity to a point P at
which the potential is 12 V?​

Answer 1

$w = {q}^{2} \div 2c \\ w = 5 \times \: {10}^{ - 3} \div 2 \times 12$

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Answer 2

Answer:

potential at infinity(Vi)= 0

potential at P(Vf)=12V

charge= 5 ×10^-3 C

work done=?

work done:

= q(Vf-Vi)

= 5× 10^-3 ( 12 - 0)

=5 × 10^-3 ( 12)

=60× 10^-3

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