Question

the workdone in moving a charge 2 ×10 to the power -9c from a point of potential -3kv to another point P is 5 × 10 to the power -5j. Find the positive at point P

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Answer 1

Question :-

The workdone in moving a charge 2 × 10 ^( -9) C from a point of potential -3k volt to another point P is 5 × 10 ^( -5) j. Find the potential at point P.

Answer :-

$\to \boxed{\sf \red{V_P = 22 \times } {10}^{3} \: volt \: } \\ \sf \: or \: \\ \to \green{ \boxed{\sf V_P = \pink{22 \: k \: volt \:} } \: }$

To Find :-

→ Potential at point P .

Explanation :-

$\bf{Given}\begin{cases}\sf{ \green{charge \: (q) =2 \times {10}^{ - 9} }\: c }\\ \sf{\purple{work\:done (w) = 5 \times {10}^{ - 5} j}}\\ \sf{ potential \:at\: point \:A \:(V_{A})}\\ \sf{V_{A } = \blue{ -3\:\times 10^{3}\: volt }}\end{cases}$

We know that :-

$\to \boxed{ \sf{ \green{Potential \: difference }= \frac{ \pink{ Work \: done}}{charge}}}$

Let potential at point P is $\sf V_P$

•°•

$\to \sf \green{ V_P -} V_A = \frac{\red{5 \times {10}^{ - 5}} }{2 \times {10}^{ - 9} } \\ \\ \to \sf \pink{V_P + 3} \times {10}^{3} = \blue{2.5 \times {10}^{4} } \\ \\ \to \sf \: V_P = 25 \times {10}^{3} - \green{3 \times {1}^{3} } \\ \\ \to \boxed{\sf V_P = \orange{22 \times {10}^{3} \: }volt \: } \\ \sf \: or \: \\ \to \boxed{\sf V_P = \red{ 22 \: k \:} volt \: }$