Answer:
Explanation:
Given that:
[tex]q_1 = + 4nC \\ \\ q_2 = -4nC \\ \\ q = 4nC[/tex]
[tex]P_1 = (0,0.04,0) m\\\\ P_2 = (0,0.05,0) m[/tex]
As [tex]q_1[/tex] [tex]\text{is nearer to }[/tex][tex]P_1 \& P_2[/tex] then [tex]q_2,q_1[/tex] [tex]\text{is positive, Thus}[/tex], [tex]E \limits^{\to}[/tex][tex]\text{ will be along the positive X-axis.}[/tex]
Recall that:
[tex]E ^{\to} = \dfrac{-dV}{dr}\implies E_y = \dfrac{-dV}{dy} \\ \\ = \int^{P_2}_{P_1}dV = -\int \limits ^{(0.05)}_{(0.04)}E_y \ dy \\ \\ \implies (V_{P_2}}-V_{(P_1)}) = \dfrac{-2q_5}{4 \pi \varepsilon _o} \int \limits ^{(0.05)}_{(0.04)} \dfrac{dy}{y^3} \\ \\ \implies (V_{P_2}}-V_{(P_1)}) = \dfrac{-2q_5}{4 \pi \varepsilon _o} \Big[ \dfrac{-1}{2y^2} \Big]^{0.05}_{0.04} \\ \\ \implies (V_{P_2}}-V_{(P_1)}) = \dfrac{q_5}{4 \pi \varepsilon _o} \Big[\dfrac{1}{y^2}\Big] ^{0.05}_{0.04}[/tex]
[tex]\\ \\ \implies (V_{P_2}}-V_{P_1}) = \dfrac{(4\times 10^{-9})(8\times 10^{-3})}{4 \pi (8.854 \times 10^{-12})} \Big [\dfrac{1}{(0.05)^2}-\dfrac{1}{(0.04)^2} \Big ]V \\ \\ (V_{P_2}}-V_{P_1}) = \mathbf{-64.71 \ volts}[/tex]
