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Question: Two charges \(5\times {{10}^{-8}}C\) and \(-3\times {{10}^{-8}}C\) are placed \(16cm\) apart. At wha...

Two charges 5×10−8C5\times {{10}^{-8}}C and −3×10−8C-3\times {{10}^{-8}}C are placed 16cm16cm apart. At what points on the line joining of the two charges will be the electric potential zero? Let us assume that the potential at infinity will be zero.

Explanation

Solution

The electric potential at a point can be found by taking the ratio of the charge at this point to the product of the four-time the value of pi, the permittivity, and the distance to this point. Apply the conditions mentioned in the question and substitute the values in it. This will help you in answering this question.

Complete step-by-step solution
Let P be the point of zero potential at a distance rr from the charge 11. Let the distance between them be given as,
d=16cmd=16cm
The electric potential at a point can be found by taking the ratio of the charge at this point to the product of the four time the value of pi, the permittivity and the distance to this point. This can be written as,
v=q4πε0dv=\dfrac{q}{4\pi {{\varepsilon }_{0}}d}
In this situation, we can write that,
v=q14πε0r+q24πε0(d−r)v=\dfrac{{{q}_{1}}}{4\pi {{\varepsilon }_{0}}r}+\dfrac{{{q}_{2}}}{4\pi {{\varepsilon }_{0}}\left( d-r \right)}
As it is already mentioned that the potential at the mentioned point is zero. Therefore we can write that,
0=q14πε0r+q24πε0(d−r)0=\dfrac{{{q}_{1}}}{4\pi {{\varepsilon }_{0}}r}+\dfrac{{{q}_{2}}}{4\pi {{\varepsilon }_{0}}\left( d-r \right)}
Simplifying this equation can be shown as,
q1r=−q2(d−r)\dfrac{{{q}_{1}}}{r}=\dfrac{-{{q}_{2}}}{\left( d-r \right)}
Substituting the values in this equation can be shown as,
5×10−8r=−3×10−8(0.16−r)\dfrac{5\times {{10}^{-8}}}{r}=\dfrac{-3\times {{10}^{-8}}}{\left( 0.16-r \right)}
From this equation, the value of rr can be found as,
r=40cmr=40cm

When the point P is outside the system, we can write the potential at this point as,
v=q14πε0r+q24πε0(r−d)v=\dfrac{{{q}_{1}}}{4\pi {{\varepsilon }_{0}}r}+\dfrac{{{q}_{2}}}{4\pi {{\varepsilon }_{0}}\left( r-d \right)}
Simplifying this equation can be shown as,
q1r=q2(r−d)\dfrac{{{q}_{1}}}{r}=\dfrac{{{q}_{2}}}{\left( r-d \right)}
Substituting the values in the equation will give,
5×10−8r=−3×10−8(r−0.16)\dfrac{5\times {{10}^{-8}}}{r}=\dfrac{-3\times {{10}^{-8}}}{\left( r-0.16 \right)}
Therefore the value of rr can be written as,
r=10cmr=10cm

Hence the answer for the question has been calculated.

Note: An electric potential is defined as the amount of work required in order to move a unit of electric charge from the reference position to a specific position in an electric field without creating an acceleration. In general, the reference points are taken as the Earth or a point at infinity.