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
 
begin mathsize 14px style F space equals space K space cross times fraction numerator q subscript 1 space q subscript 2 over denominator d squared end fraction space end style
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 begin mathsize 14px style square root of 3 squared plus 4 squared end root space equals space 5 space m end style
 
Magnitude of Force F13 between q1 = 10 μC and q3 = 2 μC is given as
 
begin mathsize 14px style F subscript 13 space equals space 9 cross times 10 to the power of 9 cross times fraction numerator 10 cross times 2 cross times 10 to the power of negative 12 end exponent over denominator 5 squared end fraction space equals space 7.2 space cross times 10 to the power of negative 3 end exponent space N end style
Force F13 in vector form = [ (7.2 × 10-3 cosθ ) begin mathsize 14px style i with hat on top end style + (7.2 × 10-3 sinθ ) begin mathsize 14px style j with hat on top end style ]  N 
where begin mathsize 14px style i with hat on top end style and begin mathsize 14px style j with hat on top end style 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)  begin mathsize 14px style i with hat on top end style + (3/5) begin mathsize 14px style j with hat on top end style ]  N = ( 5.76 begin mathsize 14px style i with hat on top end style + 4.32 begin mathsize 14px style j with hat on top end style ) × 10-3 N

Magnitude of Force F23 between q2 = -25 μC and q3 = 2 μC is given as
 
begin mathsize 14px style F subscript 23 space equals space 9 cross times 10 to the power of 9 cross times fraction numerator 25 cross times 2 cross times 10 to the power of negative 12 end exponent over denominator 5 squared end fraction space equals space 18 space cross times 10 to the power of negative 3 end exponent space N end style
Force F23 in vector form = [ 18 × 10-3 cos(180-θ ) begin mathsize 14px style i with hat on top end style + 18 × 10-3 sin(180-θ ) begin mathsize 14px style j with hat on top end style ]  N 

Force F23 = (18 × 10-3  ) [  (-4/5)  begin mathsize 14px style i with hat on top end style + (3/5) begin mathsize 14px style j with hat on top end style ]  N = ( -14.4 begin mathsize 14px style i with hat on top end style + 10.8 begin mathsize 14px style j with hat on top end style ) × 10-3 N
Net force F on q2 = 2 μC charge is
 
F = F13 + F23 = (-8.64 begin mathsize 14px style i with hat on top end style + 15.12 begin mathsize 14px style j with hat on top end style ) × 10-3 N
Magnitude of net force is given as
 
begin mathsize 14px style open vertical bar F close vertical bar space equals space square root of open parentheses 8.64 close parentheses squared plus open parentheses 15.12 close parentheses squared end root space cross times space 10 to the power of negative 3 end exponent space N space equals space 17.41 space cross times 10 to the power of negative 3 end exponent space N end style

Answered by Thiyagarajan K | 21st Jun, 2022, 02:11: PM

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