(A) Given: Charge $Q = 5\,C$,Force $F = 5000\,N$,Distance $d = 1\,cm = 10^{-2}\,m$.We know that the electric field $E = \frac{F}{Q} = \frac{5000}{5} =

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(A) Given: Charge $Q = 5\,C$,Force $F = 5000\,N$,Distance $d = 1\,cm = 10^{-2}\,m$.We know that the electric field $E = \frac{F}{Q} = \frac{5000}{5} = 1000\,N/C$.The relationship between electric field $E$,potential difference $V$,and distance $d$ is $E = \frac{V}{d}$.Therefore,$V = E \times d$.Substituting the values: $V = 1000 \times 10^{-2} = 10\,V$.Thus,the potential difference is $10\,V$.

Class 12 Physics Electric Potential and Capacitance Relation between Electric Field and Potential and Potential Gradient

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$A$ charge of $5\,C$ experiences a force of $5000\,N$ when it is kept in a uniform electric field. What is the potential difference $V$ (in volts) between two points separated by a distance of $1\,cm$?

A
$10$
B
$250$
C
$1000$
D
$2500$

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Electric Potential and Capacitance

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Relation between Electric Field and Potential and Potential Gradient

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$A$ charge of $5\,C$ experiences a force of $5000\,N$ when it is kept in a uniform electric field. What is the potential difference $V$ (in volts) between two points separated by a distance of $1\,cm$?

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