A) what is the magnitude and direction of the electric field at point A B) given

Question

A) what is the magnitude and direction of the electric field at point A

B) given that the voltage at infinity is zero, what is the voltage at A

c. explain in words what voltage or electric potential is

d. what is the magnitude and direction of the electric force on a -2.2 mC charge that is place at point A

given that the electric potential energy is zero at infinity, what is the electric potential energy of a -3.45 mC charge at point A.

diagram.

q1= -4.4 mC-----------A-------q2=7.7 mC

the distances are for the first dashed line =6.5cm and for the second dashed line=3.75 cm please explain all answers

Explanation / Answer

a) Magnitude and direction of electric field at point A:
due to -4.4 mC,
E1 = kQ/d1^2 = 9*10^9*4.4*10^-3/0.065^2
E1 = 9.37*10^9 N/C to the left
due to 7.7 mC,
E2 = kQ/d2^2 = 9*10^9*7.7*10^-3/0.0375^2 = 4.928*10^10 N/C to the left,
So net electric field at A:
5.865*10^10 N/C to the left
d) Magnitude and direction of the electric force on a -2.2 mC,
F = qE = 2.2*10^-3*5.865*10^10 to the right
F = 1.29*10^8 N. to the right
b) potential energy at point A,
PE = kQq/d1 + kQq/d2 = 9*10^9*4.4*3.45*10^-6/0.065 - 9*10^9*7.7*3.45*10^-6/0.0375 = -4273753.85 J

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