Question

Two charges, + q and - q, occupy two corners of an equilateral triangle, as shown in FIGURE 19-51. (a) If q = 1.8 mC and r = 0.50 m, find the magnitude and direction of the electric field at point A, the other vertex of the triangle. Let the direction angle be measured counterclockwise from the positive x axis. (b) What is the total electric flux through the surface indicated in the figure? (c) Explain why Gauss’s law cannot easily be used to find the magnitude of the electric field at point A.

Answer 1

We know that the electric field due to charge is given by
E = kQ/r², therefore the electric field due to charge at B and c will be
E_B = E_C = kQ/r²
E_B = E_C = 9*10⁹*(1.8*10^-3) /(0.5)² = 64.8*10⁶ N/C
Direction of field are shown in the figure
Now we will take the component of the field along the horizontal and vertical direction
Now in the vertical direction the field will cancel , therfore the net field in the vertical direction will be Zero.
Now field in the horizontal direction
E_x = E_BCos60 + E_CCos60 = 2E_BCos60
E_x  = 2* 64.8*10⁶*Cos60 = 64.8*10⁶ N/C
hence the net field at the point A will be 64.8*10⁶ N/C and the direction would be +X

(b) From gauss law we know that

where is electric flux
where Q_en is charge enclosed in the gaussian surface
Q_en= 1.8*10^-3 - 1.8*10^-3 = 0
Hence the charge enclosed is ZERO .
therefore the electric flux through the indicated surface will be zero.
(c) Because gaussian surface will not be circular therfore it will be tedious to solve integration that's why we do not use gauss law to find electric field.