There are two charges 10μc at a (0, 0) and −25μc at b (8m, 0). find the force on a charge 2μc at c (4m, 3m). express your answer as a vector.
Asked by krishankantt1061
| 21st Jun, 2022,
12:08: PM
Expert Answer:
Electric force F between two charges q1 and q2 that are separated by a distance d is given as
where K = 1/(4πεo ) = 9 × 109 N m2 C-2 is Coulomb's constant
Distance between charges q1 = 10 μC and q3 = 2 μC is 
Magnitude of Force F13 between q1 = 10 μC and q3 = 2 μC is given as
Force F13 in vector form = [ (7.2 × 10-3 cosθ ) 
+ (7.2 × 10-3 sinθ ) 
] N
where and are unit vectors along x and y direction
and θ is the angle made by direction of force F13 with x-axis.
Force F13 = (7.2 × 10-3 ) [ (4/5) + (3/5) ] N = ( 5.76 + 4.32 ) × 10-3 N
Magnitude of Force F23 between q2 = -25 μC and q3 = 2 μC is given as
Force F23 in vector form = [ 18 × 10-3 cos(180-θ ) + 18 × 10-3 sin(180-θ ) ] N
Force F23 = (18 × 10-3 ) [ (-4/5) + (3/5) ] N = ( -14.4 + 10.8 ) × 10-3 N
Net force F on q2 = 2 μC charge is
F = F13 + F23 = (-8.64 + 15.12 ) × 10-3 N
Magnitude of net force is given as
+ (7.2 × 10-3 sinθ ) ] N and are unit vectors along x and y direction + (3/5) ] N = ( 5.76 + 4.32 ) × 10-3 NMagnitude of Force F23 between q2 = -25 μC and q3 = 2 μC is given as
Force F23 in vector form = [ 18 × 10-3 cos(180-θ ) + 18 × 10-3 sin(180-θ ) ] N
+ 18 × 10-3 sin(180-θ ) ] N Force F23 = (18 × 10-3 ) [ (-4/5) + (3/5) ] N = ( -14.4 + 10.8 ) × 10-3 N
+ (3/5) ] N = ( -14.4 + 10.8 ) × 10-3 NNet force F on q2 = 2 μC charge is
F = F13 + F23 = (-8.64 + 15.12 ) × 10-3 N
+ 15.12 ) × 10-3 NMagnitude of net force is given as
Answered by Thiyagarajan K
| 21st Jun, 2022,
02:11: PM
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