Answer to Question #172656 in Physics for hazal

Question

Calculate the electric field (magnitude and direction) at the upper right corner of a square 1.22 m

m on a side if the other three corners are occupied by 3.00×10⁻⁶ C

C charges. Assume that the positive x-axis is directed to the right.

∘

?∘ counterclockwise from the x-direction

Answer 1

Let's consider the square ABCD. The three corners of the square (B, C and D) are occupied by the charges of magnitude $3.0\cdot10^{-6}\ C$. We are searching for the magnitude and direction of the electric field at corner A. Let's find the magnitude of the electric fields at B, C and D:

$E_B=\dfrac{kQ_B}{r^2}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot3.0\cdot10^{-6}\ C}{(1.22\ m)^2}=18140\ \dfrac{V}{m},$ $E_C=\dfrac{kQ_C}{(r\sqrt{2})^2}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot3.0\cdot10^{-6}\ C}{(1.22\ m\cdot\sqrt{2})^2}=9070\ \dfrac{V}{m},$ $E_D=\dfrac{kQ_D}{r^2}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot3.0\cdot10^{-6}\ C}{(1.22\ m)^2}=18140\ \dfrac{V}{m}.$

Finally, we can find the net electric field at A:

$E_A=E_Bcos\theta+E_C+E_Dcos\theta.$

We can find the angle from the geometry:

$cos\theta=\dfrac{r}{r\sqrt{2}},$ $\theta=cos^{-1}(\dfrac{1}{\sqrt{2}})=45^{\circ}.$

Finally, we get:

$E_A=18140\ \dfrac{V}{m}\cdot cos45^{\circ}+9070\ \dfrac{V}{m}+18140\ \dfrac{V}{m}\cdot cos45^{\circ}=34724\ \dfrac{V}{m}.$

Answer:

$E_A=34724\ \dfrac{V}{m}.$

$\theta=45^{\circ}.$