Question

What is the magnitude and direction of the electric field at the point P in the configuration of charges pictured? The distance between each charge and point P is 0.1 m, each charge is q 2 mC, and there is a 120 degree angle between the lines connecting the charges and P. 120

Answer 1

pt. rightarrow point Electric field on pt. p due to charge at pt. A is E vector_A = 1/4 pi epsilon_0 q_A/r^2_A r^^_A (i) and due to charge at pt. B is E vector_B = 1/4 pi epsilon_0 q_B/r^2_B r^^_B (ii) r^^_A and R^^_B are unit vectors along the directions of E vector_A & E vector_B respectively eqns. (i) and (ii) are from Coulomb�s law and q_A = q_B = 2 mc = 2 times 10^-3 c So, r_A = r_B = 0.1 m E vector_A = 9 times 10^3 times 2 times 10^-3/(0.1)^2 r^^_A Because 1/4 pi epsilon_0 = 9 times 10^3 Nm^2/c^2 Coulomb�s constant = 1.8 times 10^3 F_A N/c Similarly, E vector_B = 1.8 times 10^10^9 r^^-B N/c Now, since Electric field is vector quantity, so the resultant of F vector_A and E vector_B has a magnitude Vertical-bar E vector Vertical-bar = squareroot Vertical-bar E vector_A Vertical-bar ^2 + Vertical-bar E vector_B Vertical-bar ^2 + 2 E vector_A middot E vector_B

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