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Question: We are given that charges \({{q}_{1}}=2C\) and \({{q}_{2}}=-3C\) are placed at \((0,0)m\) and \((100...

We are given that charges q1=2C{{q}_{1}}=2C and q2=−3C{{q}_{2}}=-3C are placed at (0,0)m(0,0)m and (100,0)m(100,0)m points respectively. At what points on the X-axis is the electric potential zero?

Explanation

Solution

The electric potential difference is the work done to move a charge from one point to the other in the presence of an electric field. The point where the potential due to both charges is equal is the point where the potential is zero. Calculate the potential due to the charges and equate them to find the point.
Formulas used:
V=q4πε0rV=\dfrac{q}{4\pi {{\varepsilon }_{0}}r}
V1−V2=0{{V}_{1}}-{{V}_{2}}=0

Complete step-by-step solution:
The electric potential at a point is defined as the work done to bring a unit charge at that point. It’s SI unit is Volt (V). Therefore,
V=q4πε0rV=\dfrac{q}{4\pi {{\varepsilon }_{0}}r} -------- (1)
Here, VV is the potential at a point
qq is the magnitude of charge
rr is the distance of point to the charge
The point on the x-axis where the potential is zero is the point where potential due to both charges is equal. Therefore,
V1−V2=0{{V}_{1}}-{{V}_{2}}=0 ----------- (2)
Given, two charges q1=2C{{q}_{1}}=2C and q2=−3C{{q}_{2}}=-3C are placed at (0,0)m(0,0)m and (100,0)m(100,0)m respectively.
Using eq (1) and eq (2), we have,
24πε0x=−34πε0(100−x) ⇒2x=−3100−x ⇒2(100−x)=−3x ⇒200−2x=−3x ⇒200=−x ∴x=−200m \begin{aligned} & \dfrac{2}{4\pi {{\varepsilon }_{0}}x}=\dfrac{-3}{4\pi {{\varepsilon }_{0}}(100-x)} \\\ & \Rightarrow \dfrac{2}{x}=\dfrac{-3}{100-x} \\\ & \Rightarrow 2(100-x)=-3x \\\ & \Rightarrow 200-2x=-3x \\\ & \Rightarrow 200=-x \\\ & \therefore x=-200m \\\ \end{aligned}
The point on the x-axis where the potential due to both charges is zero is −200m-200m.
Therefore, the point at which potential due to both charges is zero is (−200,0)m(-200,0)m

Note: The potential energy due to a negative charge is negative because work is done in the direction of the field while potential due to positive charge is positive because work done is opposite to the direction of electric field. The work done to bring a unit charge from infinity to a point is the electric field. It is not necessary that the potential is zero where the electric field is zero.