Answer to Question #177107 in Electricity and Magnetism for Deep

Question

Consider four positive charges with charge Q arranged in a square with side-length d, centred about the origin in the x − y plane. (a) Sketch the electric field of the charge configuration, using field lines. Describe how you constructed your sketch, and list some key features of charges and electric field lines that your sketch demonstrates. (b) Derive an expression for the electric potential V (x, y) in the x − y plane (c) Describe, using concepts and terminology used in PHYS1002, the difference between electric potential and potential energy. (d) Derive an expression for the potential energy U of the charge configuration, and show that it is given by U = (4 + √ 2) 4π0 Q2 d . (e) The charges are all released simultaneously from rest, and accelerate away from each other due to Coulomb repulsion. If Q = 22.4 µC, d = 1.50 mm and the mass of each is 1.00 g, calculate their speed when they are very far apart. Include appropriate significant figures in your answer.

Answer 1

The electric field of the line due to +q charge

a) The center electric field be zero

b) Electric potential V(x) at the center

$d^2+d^2=x^2$

$2d^2=x^2$

$\frac{x}{2}=\frac{\sqrt{2}d}{2}$

Total path =$V_1+V_2+V_3+V_4$

$V_{Total}=4V_1$

$V_1=\frac{kq}{0.5x}= \frac{kq}{ \frac{\sqrt{2}d}{2}}= \frac{2kq}{\sqrt{2}d}$

$V_{Total}=4 \times \frac{2kq}{\sqrt{2}d}$

$V_{Total}=\frac{8kq}{\sqrt{2}d}$

$V_{Total}= \frac{4\sqrt{2} kq}{d}$

$V_{Total}= \frac{4\sqrt{2} q}{4 \pi \epsilon_0 d}$

$V_{Total}= \frac{\sqrt{2} q}{\pi \epsilon_0 d}$

c) Difference between electric potential and potential energy

Electric Potential

An electric path in an electric field is the amount of work done to bring the unit position charge from infinity to that point.

Potential energy

Electric potential energy is the energy that is needed to move a charge partially against the electric field.

d) $U= \frac{1}{4\pi \epsilon_0} \sum \frac{q_iq_j}{r_{ij}}$

$U= \frac{1}{4\pi \epsilon_0} (\frac{q_1q_2}{r_{12}}+\frac{q_2q_3}{r_{23}}+\frac{q_3q_4}{r_{34}}+\frac{q_1q_3}{r_{13}}+\frac{q_1q_4}{r_{14}}+\frac{q_2q_4}{r_{24}})$

$U= \frac{1}{4\pi \epsilon_0} (\frac{q^2}{d}+\frac{q^2}{d}+\frac{q^2}{d}+\frac{q^2}{ \sqrt{2}d}+\frac{q^2}{d}+\frac{q^2}{ \sqrt{2}d})$

$U= \frac{1}{4\pi \epsilon_0} (4\frac{q^2}{d}+2\frac{q^2}{ \sqrt{2}d})$

$U= \frac{1}{4\pi \epsilon_0} (\frac{4q^2}{d}+\frac{\sqrt{2}q^2}{ d})$

$U= \frac{1}{4\pi \epsilon_0} \frac{q^2}{d}(4+\sqrt{2})$

e)$F_1=\frac{kQ^2}{d^2}$

$F_4=\frac{kQ^2}{d}$

$F_3=\frac{kQ^2}{({\sqrt{2}d})^2}$

$F_{net}=F_3+$ Resultant of F₁ and F₄

$F_{net}=\frac{kQ^2}{2d^2}+\sqrt{(\frac{kQ^2}{d^2})^2+(\frac{kQ^2}{d^2})^2+(\frac{kQ^2}{d^2})^2cos90}$

$F_{net}=\frac{kQ^2}{2d^2}+\frac{kQ^2}{d^2}\sqrt{2}$

$F_{net}=\frac{kQ^2}{d^2}(\frac{1}{2}+\sqrt{2})$

Write equation of motion

$F_{net}=ma$

$\frac{kQ^2}{d^2}(\frac{1}{2}+\sqrt{2})=m \frac{Udu}{dx}$

$\frac{9 \times 10^9 \times (22.4 \times10^{-6})^2}{(1.5 \times10^{-3})^2}(\frac{1}{2}+\sqrt{2})=10^{-3}\frac{Udu}{dx}$

$3841.9031 \times10^6=\frac{Udu}{dx}$

$\int_0^{v}{Udu}= \int_0^{\infin}3841.9031 \times10^6dx$

Initial charge is at rest

$\frac{v^2}{2}=\infin$

$v=\infin$