System of Electric Charges
Please see the attached file.
3) A particle carrying charge qA = -8 × 10^-4 C is placed at point A and a particle carrying charge qB = -2 × 10^-4 C is placed at point B.
a) Find the magnitude and direction of the electric field at point C.
b) Find the electric potential at point C.
c) Find the direction and magnitude of the electrostatic force acting on the particle carrying charge qC = -3 × 10^-5 C placed at point C.
d) How much work would one need to do to move particle qC from its location at C to infinity?
e) Is there a point along the line joining charges qA and qB where if you placed charge qC it will experience no acceleration? If so, where? Specify the distance to qA.
https://brainmass.com/physics/system-work/system-electric-charges-174309
SOLUTION This solution is FREE courtesy of BrainMass!
Please refer to the attachment.
C D
A 2m 4m B
x 6-x
qA = -8 x 10-4 C
qB = -2 x 10-4 C
qC = -3 x 10-5 C
a) Magnitude of electric field at C due to qA = EA = (1/4Πε0)qA/rAC2 = 9x109x(8 x 10-4)/22 = 18x105 N/C. Direction of EA is towards left because a unit +ve charge placed at C will be attracted by charge at A towards left.
Magnitude of electric field at C due to qB = EB = (1/4Πε0)qB/rBC2 = 9x109x(2 x 10-4)/42 = 1.125x105 N/C. Direction of EB is towards right because a unit +ve charge placed at C will be attracted by charge at B towards right.
Net electric field at C = EC = EA - EB = 18x105 - 1.125x105 = 16.88x105 N/C towards left
b) Electric potential at C due to qA = VA = (1/4Πε0)qA/rAC = 9x109x(-8 x 10-4)/2 = -36x105 V.
Electric potential at C due to qB = VB = (1/4Πε0)qB/rBC = 9x109x(-2 x 10-4)/4 = - 4.5x105 V.
Net electric potential at C = VC = VA + VB = -36x105 - 4.5x105 = - 40.5x105 J/C (Or Volts)
c) Electrostatic force on a charge q placed at a point where electric field is E is given by qE. Hence, magnitude of force on charge qC = qC x EC = 3x10-5x16.88x105 = 50.64 N. As a negative charge experiences force in the direction opposite to the direction of E, the negative charge at C will experience force towards right (as net electric field at C is towards left).
d) Electric potential at a point is defined as the work done is moving a unit positive charge from infinity to that point. Hence, work done in moving a unit positive charge from a given point to infinity must be negative of the potential at the point. Hence, work done in moving unit positive charge from C to infinity = - VC = 40.5x105 J/C
Work done in moving qC from C to infinity = - 3x10-5x40.5x105 = - 121.5 J
e) For qC to experience zero acceleration, it must be placed at a point where electric field intensity is zero. Let D be such a point at a distance x from A (distance from B = 6-x). Electric field at D due to both the charges must be equal and opposite. Hence,
9x109x(8 x 10-4)/x2 = 9x109x(2 x 10-4)/(6-x)2
Or 4 /x2 = 1/(6-x)2 or 4(6-x)2 = x2 or 4(36 + x2 - 12x) = x2
Or 3x2 - 48x + 144 = 0 Solving we get x = 4m or 12m from charge A.
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