Question

Point charges q₁ = +2.00 µC and q₂ = -2.00 µC are placed at adjacent corners of a square for which the length of each side is 3.00 cm. Point a is at the center of the square, and point b is at the empty corner closest to q₂. Take the electric potential to be zero at a distance far from both charges.

(a) What is the electric potential at point a due to q₁ and q₂?

(b) What is the electric potential at point b?

(c) A point charge q₃ = -5.00 µC moves from point a to point b. How much work is done on q₃ by the electric forces exerted by q₁ and q₂? Is this work positive or negative?

Short answer

1. The electric potential at a point due to q₁ and q₂ is Zero.
2. The electric potential at b is -175 KV.
3. the work done in moving a charge from a to be is -0.875 Joules

Step-by-step solution

Given Data and concept

q₁ = 2.0010^-6 C

q₂ = - 2.0010^-6 C

q₃ = -5.0010^-6 C

The electric potential at point r due to point charge q can be written as:

$\mathit{V}\mathbf{=}\frac{\mathbf{K}\mathbf{q}}{\mathbf{r}}$

Here V is the potential.

Electric Potential due to q1 and q2

(a)

Electric Potential due both charges at a point is given by,

$$\begin{gathered} V_a=\frac{k{q}_1}{d}+\frac{k{q}_2}{d} \\ =\frac{k}{d}\left(2.00\mu C-2.00\mu c\right) \\ =0 \end{gathered}$$

Therefore, the electric potential at a point due to q₁ and q₂ is Zero.

Electric Potential at b

(b)

The electric potential at b is given by,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{b}}=\frac{{\mathrm{kq}}_1}{{\mathrm{d}}_1}+\frac{{\mathrm{kq}}_2}{{\mathrm{d}}_2} \\ =\mathrm{k}\left(\frac{{\mathrm{q}}_1}{{\mathrm{d}}_1}+\frac{{\mathrm{q}}_2}{{\mathrm{d}}_2}\right) \\ =9\times {10}^9\left(\frac{2.0\times {10}^{-6}}{0.0424}-\frac{2.0\times {10}^{-6}}{0.03}\right) \\ =-1.75\mathrm{KV} \end{gathered}$$

Therefore, the electric potential at b is -175 KV.

Work done on q3

(c)

The work done on charge q₃ from a to b is given by,

$$\begin{gathered} \mathrm{W}={\mathrm{q}}_3\left({\mathrm{V}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{b}}\right) \\ =\left(-5\times {10}^{-6}\right)\left(175\times {10}^3-0\right) \\ =-0.875\mathrm{J} \end{gathered}$$

Therefore, the work done moving a charge from a to be is -0.875 Joules.