a) What is the electric field (magnitude and direction) at the center of the squ

Question

a) What is the electric field (magnitude and direction) at the center of the square?

b) what is the electric potential at the center of the square?

c) how much energy would it take to move a 2 micro Coulomb charge from a very long distance to the center of the square?

Don't mind the work I began to do for part a. It may not be correct!

2. Consider four charges placed at the comer ofa square of side 10em as show. a) What is the electric field (magnitude and direction) at the center of the square? b) What is the electric Potential at the center of the square? c) How much energy would it take to move a2HC charge from a very long distance to the center of the square?

Explanation / Answer

a) The -1 mu C charges on the two diagnol points of the square would cancel each other's electric field at the centre of the suare through sym,metry
the -1.5mu C charges would have electric field in same direction , so would algebraically add up ( in the direction from +1.5 muC to -1.5mu C charge)

so, E at centre of square = kq/r^2 + kq'/r^2 = k(1.5 + 1.5)*10^-6/(0.1COS(45))^2 = 8.98*10^9*3*10^-6/0.01*2 = 1347000 V/m in the direction joining 1.5mu C to -1.5 mu C

b) Electric potential is a scalar so we can addup potential due to all the charges indivdually at the centre of the square
so potential due to a point charge q at a distacne r from the point, V = kq/r
so, net potential at the centre of the square , V = k(1.5 - 1 - 1.5 - 1)*10^-6/0.1cos(45) = 8.98*10^9*(-2)*10^-6/0.1cos(45) = -2539927.7558 V

c) Energy required = PE at infinity - PE at the centre of the square
PE for a charge q at a point in space with potential V = qV
so, energy required = q(dV) = q(0 - V) = -qV ( assuming potential at infinity be 0)
E = 2*10^-6*(-2539927.7558 ) = -0.5079 J

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