Two tiny particles having charges $+ 20.0 \mu \mathrm { C }$ and $- 8.00 \mu \mathrm { C }$ are separated by a distance of $20.0 \mathrm {~cm}$. What are the magnitude and direction of electric field midway between these two charges? $( k = 1 / 4$ $\pi \varepsilon _ { 0 } = 9.0 \times 10 ^ { 9 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 }$ )

Question

Quizplus · Accepted Answer

The electric field at the midpoint due to each charge is calculated using the formula $E = k \frac{|q|}{r^2}$, where $r = 0.1$ m is the distance from each charge to the midpoint. For the $+20.0 \mu C$ charge, $E_1 = 9.0 \times 10^9 \times \frac{20.0 \times 10^{-6}}{(0.1)^2}$, and for the $-8.00 \mu C$ charge, $E_2 = 9.0 \times 10^9 \times \frac{8.00 \times 10^{-6}}{(0.1)^2}$. The net electric field is the difference between these two fields because they are in opposite directions due to opposite charges. The direction of the net field is towards the negative charge because the electric field points away from positive charges and towards negative charges.

Two Tiny Particles Having Charges +20.0μC+ 20.0 \mu \mathrm { C }+20.0μC

Two Tiny Particles Having Charges $+ 20.0 \mu \mathrm { C }$