Question

When a particle of charge 10 uC is brought from infinity to point in the electric field, 10mj work is done by the external force. What is the potential at that point? ...........here why we need to multiply 10^6×10 and 10^3×10​

Answer 1

$v = \frac{w}{q}$

v = potential difference

w = work done

q = charge

$1\mu \: = {10}^{ - 6} \\ 1m = {10}^{ - 3}$

so,

$v = \frac{10 \times {10}^{ - 6} }{10 \times {10}^{ - 3} } \\ v = {10}^{ - 3} = 1m$

ans = potential difference is 1 mV

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