4. \"Charge A\" is + 2 x 106C and is at (o,o) on thex-y axes. \"Charge B\" has a

Question

4. "Charge A" is + 2 x 106C and is at (o,o) on thex-y axes. "Charge B" has a magnitude of -3 x 106 C and is at (0, 1). "Charge C" is +1x10 C and is located at (1, 2). All dimensions are in meters. Calculate the force exerted on charge C by the presence of the other two charges. (Find magnitude and give direction relative to thel X-axis) What is the magnitude and direction ofthe electric field at a distance of 5 meters from a positive chargeof 3 x 10"N/C? 5. 6. Answer the same question as #5 fora negative charge.

Explanation / Answer

force between two charged particles = k * q1 * q2 / r^2

where k = 9 * 10^9 Nm^2/C^2

force between particle A and C

F1 = 9 * 10^9 * 2 * 10^-6 * 1 * 10^-6 / 5

F1 = 0.0036N

force between particle B and C

F2 = 9 * 10^9 * (-3) * 10^-6 * 1 * 10^-6 / 1

F2 = -0.027N

angle between forces = 180-45

angle between forces = 135 degree

resultant = sqrt(F1^2 + F2^2 + 2F1 * F2 cos(135))

resultant = sqrt(0.0036^2 + 0.027^2 + 2*0.0036 * 0.027* cos(135))

resultant force = 0.0246 N

angle = tan^-1(F2 * sin(135) / F1 + F2 cos(135))

angle = tan^-1(0.027 * sin(135) / (0.0036 + 0.027 cos(135)))

angle = -50.94 degree

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