$(a)$ Calculate the potential at a point $P$ due to a charge of $4 \times 10^{-7} \; C$ located $9 \; cm$ away.
$(b)$ Hence,obtain the work done in bringing a charge of $2 \times 10^{-9} \; C$ from infinity to the point $P$. Does the answer depend on the path along which the charge is brought?
$(b)$ Hence,obtain the work done in bringing a charge of $2 \times 10^{-9} \; C$ from infinity to the point $P$. Does the answer depend on the path along which the charge is brought?
(N/A) The electric potential $V$ at a distance $r$ due to a charge $Q$ is given by $V = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}$.
Given $Q = 4 \times 10^{-7} \; C$ and $r = 9 \; cm = 0.09 \; m$.
$V = (9 \times 10^{9} \; N \cdot m^{2} \cdot C^{-2}) \times \frac{4 \times 10^{-7} \; C}{0.09 \; m} = 4 \times 10^{4} \; V$.
$(b)$ The work done $W$ in bringing a charge $q$ from infinity to a point $P$ is $W = qV$.
Given $q = 2 \times 10^{-9} \; C$.
$W = (2 \times 10^{-9} \; C) \times (4 \times 10^{4} \; V) = 8 \times 10^{-5} \; J$.
No,the work done is path-independent because the electrostatic force is a conservative force. The work done depends only on the initial and final positions.
Given $Q = 4 \times 10^{-7} \; C$ and $r = 9 \; cm = 0.09 \; m$.
$V = (9 \times 10^{9} \; N \cdot m^{2} \cdot C^{-2}) \times \frac{4 \times 10^{-7} \; C}{0.09 \; m} = 4 \times 10^{4} \; V$.
$(b)$ The work done $W$ in bringing a charge $q$ from infinity to a point $P$ is $W = qV$.
Given $q = 2 \times 10^{-9} \; C$.
$W = (2 \times 10^{-9} \; C) \times (4 \times 10^{4} \; V) = 8 \times 10^{-5} \; J$.
No,the work done is path-independent because the electrostatic force is a conservative force. The work done depends only on the initial and final positions.
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