Problem 1 (25 points) 2Q 4Q 3Q At each corner of a square of side L there are fo

Question

Problem 1 (25 points) 2Q 4Q 3Q At each corner of a square of side L there are four point charges. Determine the force (in terms of unit vectors i and j) on the charge 2Q in terms ofk,Q, and L. Do not find the force in terms of magnitude and direction. Simplify your expressions as much as possible (all numerical expressions should be converted to real numbers). Problem 2 (25 points) For the figure above, find the electric field (in terms of unit vectors i and j ) at the center of the square in terms of k,Q, and IL. Do not find the electric field in terms of magnitude and direction. Simplify your expressions as much as possible (all numerical expressions should be converted to real numbers). Problem 3 (25 points) For the figure above (a) find the electrie potential energy for the array of charges in terms of k ,Q , and L. (b) find the electric potential (voltage) at the center of the square in terms of k,Q and L simplify your expressions as much as possible (all numerical expressions should be converted to real numbers)

Explanation / Answer

2. for the given charge distribution

consider an axis to the right be x and to the top be y

then considering the i and j unit vectors

electric field at the midpoint of the squar ein fector notation is

E = 2kQ(i - j)cos(45)/L^2 + 4kQ(-i - j)cos(45)/L^2 + 6kQ(-i + j)cos(45)/L^2 + 8kQ(i + j)cos(45)/L^2

E = 2kQ(i - j -2i - 2j -3i + 3j + 4i + 4j)cos(45)/L^2

E = 2kQ(4j)cos(45)/L^2 = 8kQj/sqroot(2)L^2

3. electric potnetial energy of the system of charges in the figure can be given by

U = kQ*2Q/L + k*Q*3Q/L*sqrt(2) + kQ*4Q/L + k*2Q*3Q/L + k*2Q*4Q/L*sqrt(2) + k*3Q*4Q/L

U = 24kQ^2/L + 11k*Q^2/L*sqrt(2)

U = kQ^2(24 + 11/sqrt(2))/L = 31.77817*kQ^2/L

electric potential at the center of the square is given by

V = k(Q + 2Q + 3Q + 4Q)/sqroot(2)L = 10kQ/L*sqrt(2)

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