A -9.0 times 10^-5-C charge is fixed on the x-axis at 8 m, and a +1.1 times 10^-

Question

A -9.0 times 10^-5-C charge is fixed on the x-axis at 8 m, and a +1.1 times 10^-5- C charge is fixed on the y-axis at -5 m as shown below. The electric force constant k is 8.99 times 10^9 Nm^2/C^2. Find the magnitude and direction of the resultant electric field at the origin (0m, 0m). Find the electric potential at point the origin. A -4 times 10 -C charge is brought from a very distant point by an external force and placed at the origin. Determine the magnitude and direction of the electric force on the charge.

Explanation / Answer

Let E1 be the electric field at the origin due to 1.1*10^-5 C

E1 = k*q1/x1^2 = (8.99*10^9*9*10^-5)/(8^2) = 12642.18 N/C along +X-axis

similargy due to 1.1*10^-5 C is E2 = k*q2/y^2 = (8.99*10^9*1.1*10^-5)/(5^2) = 3955.59 N/C along +Y-axis

Net electric field is E = sqrt(E1^2+E2^2) =sqrt(12642.18^2+3955.59^2) = 1.32*10^4 N/C

direction is theta = tan^(-1)(E2/E1) = tan^(-1)(3955.59/12642.18) = 17.4 degrees above the +X-axis

ii) Electic potential is V = V1+V2

V1 = k*q1/x = (-8.99*10^9*9*10^-5)/(8) = -1.011*10^5 V

V2 = k*q2/y = (8.99*10^9*1.1*10^-5)/5 = 1.9778*10^4 V

V = V1+V2 = (-1.011+0.19778)*10^5 = -8.1322*10^4 V

3) F = q*E = -4*10^-6*1.32*10^4 = -0.0528 N opposite to the direction of E

i.e 17.4 degrees below the negative X-axis

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