Question

If 120j of work is done in car to carrying a charge of 6c from a place where the potential is 10 volt to another place where the potential is v find v.

Answer 1

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Given work done = 120 J

charge q = 6 C

from potential V1 = 10 Volts to V2 = V Volts

so,

$\sf \: W=q△V \\ \\ \dashrightarrow \sf \: W=q(V2 −V1) \\ \\ \dashrightarrow \sf \: W=6 [V2 − 10] \\ \\ \sf \dashrightarrow \cancel \frac{120}{6} =V2 - 10 \\ \\ \dashrightarrow\sf \: 20 = V2 - 10 \\ \\ \dashrightarrow \sf \: V2 = 20 + 10 \\ \\ \dashrightarrow\sf \boxed{ \sf \red{V2 = 30 \: volts}}$

Answer 2

Given:

120J of work is done in car to carrying a charge of 6C from a place where the potential is 10 volt to another place where the potential is V.

To find:

Value of V ?

Calculation:

We know that :

• Potential difference is defined as the work done to transfer unit positive charge from one place to another in a uniform electric field.

$PD = \dfrac{work}{charge}$

$\implies \: V-10 = \dfrac{120}{6}$

$\implies \: V-10 = 20$

$\implies \: V = 10 +20$

$\implies \: V = 30 \: Volts$

So , value of V is 30 Volts.