Calculate the potential at a point \(P\) due to a charge of \(4\times 10^{-7}C\) located \(9\text{ cm}\) away. 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?
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Calculate the potential at a point \(P\) due to a charge of \(4\times 10^{-7}C\) located \(9\text{ cm}\) away. 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?
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Solution
Given :
\(Q=4\times 10^{-7}C,r=0.09\;m\)
For source charge, Potential at \(P\) due to \(Q\) charge,
\(V_P=\left(\frac 1{4\pi \varepsilon _0}\right)\left(\frac Q r\right)\)
\(=\frac{\left(9\times 10^9\right)\left(4\times 10^{-7}\right)}{0.09}\)
\(=4\times 10^4\;V\)
Given data,
Test charge, \(q=2\times 10^{-9}C\)
Potential at point \(P=\) \(4\times 10^4\;V\)
Potential at a point (test charge) is the amount of work done to bring unit charge from infinity to that point.
\(W_{{\infty}P}=qV_p\)
\(=2\times 10^{-9}\times 4\times 10^4\)
\(=8\times 10^{-5}\;J\)
No, this work done does not depend on the path followed. The reason is that this work is against electrostatic force which is a conservative force.
\(Q=4\times 10^{-7}C,r=0.09\;m\)
For source charge, Potential at \(P\) due to \(Q\) charge,
\(V_P=\left(\frac 1{4\pi \varepsilon _0}\right)\left(\frac Q r\right)\)
\(=\frac{\left(9\times 10^9\right)\left(4\times 10^{-7}\right)}{0.09}\)
\(=4\times 10^4\;V\)
Given data,
Test charge, \(q=2\times 10^{-9}C\)
Potential at point \(P=\) \(4\times 10^4\;V\)
Potential at a point (test charge) is the amount of work done to bring unit charge from infinity to that point.
\(W_{{\infty}P}=qV_p\)
\(=2\times 10^{-9}\times 4\times 10^4\)
\(=8\times 10^{-5}\;J\)
No, this work done does not depend on the path followed. The reason is that this work is against electrostatic force which is a conservative force.
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