Four point charges are located at thevertices of a diamond, with the centeratthe

Question

Four point charges are located at thevertices of a diamond, with the centeratthe origin. The top charge is -2 C, the charge onthe right side is 4C, the charge at the bottom of the diamondis -2C, and the charge on the left side of the diamond is3C. All charges are a distance of 1 meter away from theorigin at the center.

a. Calculate theElectric Field vector atthe position of the 4 C charge that is due to

the other three charges.

b. Calculate theElectric Force vector onthe 4 C charge due to theother three

charges.

c. Calculate theEnergy necessary to bring inthe 4 C charge, presuming theother

three are already inplace.

d. What is theElectric Potential (Voltage) at the pointP due to the fourcharges?

Assume that the point P isat the coordinate (1, 1).

Explanation / Answer

a.The electric field vector at the position of the 4 Ccharge that is due to the other three charges is E = (Ex2 +Ey2)1/2 where Ex = E1x + E2x +E3x = E1 * cos1 +E2 * cos2 + E3 *cos3 and Ey = E1y + E2y +E3y = E1 * sin1 +E2 * sin2 + E3 *sin3 where E1 = k *(q1/r12) k = (1/4o) = 9 * 109Nm2/C2,q1 = 2 C = 2 *10-6 C and r12 = [(1 -0)2 + (0 - 1)2] = 2 m2 similarly,E2 = k *(q2/r22) q2 = 3 C = 3 * 10-6 C andr22 = [(1 - (-1))2 + (0 -1)2] = 5 m2 and E3 = k *(q3/r32) q3 = -2 C = -2 * 10-6 Cand r32 = [(0 - 1)2 + (-1 -0)2] = 2 m2 The angles are tan1 = (0 - 1/1 - 0) = -1 or 1 = tan-1(-1) = -45o similarly,tan2 = (0 - 0/1 - (-1)) = 0 or 2 = tan-1(0) =0o and tan3 = (-1 - 0/0 - 1) = 1 or 3 = tan-1(1) =45o Therefore,the electric field vector at the position ofthe 4 C charge is E = iEx + jEy The angle made by the resultant electric field is tan = (Ey/Ex) or =tan-1(Ey/Ex) b.The electric force vector at the position of the 4C charge is F = i(Ex/q) + j(Ey/q) q = 4 C = 4 * 10-6 C c.The energy increase to bring the 4 C charge is U = F * d where d = 1 m d.The electric potential at the point P due to the fourcharges is V = V1 + V2 + V3 +V4 where V1 = k * (q1/r1) r1 = [(0 - (-1))2 + (1 -1)2]1/2 = 1 m V2 = k * (q2/r2) r2 = [(-1 - (-1))2 + (0 -1)2]1/2 = 1 m V3 = k * (q3/r3) r3 = [(0 - (-1))2 + (-1 -1)2]1/2 = 5 m V4 = k * (q4/r4) r4 = [(1 - (-1))2 + (0 -1)2]1/2 = 5 m similarly,tan2 = (0 - 0/1 - (-1)) = 0 or 2 = tan-1(0) =0o and tan3 = (-1 - 0/0 - 1) = 1 or 3 = tan-1(1) =45o Therefore,the electric field vector at the position ofthe 4 C charge is E = iEx + jEy The angle made by the resultant electric field is tan = (Ey/Ex) or =tan-1(Ey/Ex) b.The electric force vector at the position of the 4C charge is F = i(Ex/q) + j(Ey/q) q = 4 C = 4 * 10-6 C c.The energy increase to bring the 4 C charge is U = F * d where d = 1 m d.The electric potential at the point P due to the fourcharges is V = V1 + V2 + V3 +V4 where V1 = k * (q1/r1) r1 = [(0 - (-1))2 + (1 -1)2]1/2 = 1 m V2 = k * (q2/r2) r2 = [(-1 - (-1))2 + (0 -1)2]1/2 = 1 m V3 = k * (q3/r3) r3 = [(0 - (-1))2 + (-1 -1)2]1/2 = 5 m V4 = k * (q4/r4) r4 = [(1 - (-1))2 + (0 -1)2]1/2 = 5 m where V1 = k * (q1/r1) r1 = [(0 - (-1))2 + (1 -1)2]1/2 = 1 m V2 = k * (q2/r2) r2 = [(-1 - (-1))2 + (0 -1)2]1/2 = 1 m V3 = k * (q3/r3) r3 = [(0 - (-1))2 + (-1 -1)2]1/2 = 5 m V4 = k * (q4/r4) r4 = [(1 - (-1))2 + (0 -1)2]1/2 = 5 m V3 = k * (q3/r3) r3 = [(0 - (-1))2 + (-1 -1)2]1/2 = 5 m V4 = k * (q4/r4) r4 = [(1 - (-1))2 + (0 -1)2]1/2 = 5 m V4 = k * (q4/r4) r4 = [(1 - (-1))2 + (0 -1)2]1/2 = 5 m

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