(a) Calculate potential at a point P due to a charge of 4 × 10^{-7} C located 9 cm away. (b) Hence

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2. Electric Potential and Capacitance

easy

$(a)$ 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?

Option A

Option B

Option C

Option D

Solution

$(a)$ $V=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}=9 \times 10^{9} \,Nm ^{2}\, C ^{-2} \times \frac{4 \times 10^{-7} \,C }{0.09 \,m }$

$=4 \times 10^{4}\, V$

$(b)$ $W=q V=2 \times 10^{-9} \,C \times 4 \times 10^{4}\, V$

$=8 \times 10^{-5} \,J$

No. work done will be path independent. Any arbitrary infinitestmal path can be resolved into two perpendicular displacements: One along $r$ and another perpendicular to $r$. The work done corresponding to the later will be zero.

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$(a)$ 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?

Solution

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$(a)$ 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?