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A point charge 50mC is located in the XY plane at the point of position vector . What is the electric field at the point of position vector 
  • a)
    1200 V/m
  • b)
    0.04 V/m 
  • c)
    900 V/m
  • d)
    45000 V/m
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A point charge 50mC is located in the XY plane at the point of positio...
Field intensity is proportional to the inverse of the square of the distance separating the point charges:
F = k*q1*q2/d^2
Since k, q1, and q2 are assumed to be constant, their product can be combined to form a single constant:
K = k*q1*q2, so
F = K/d^2 (and by algebraic manipulation we have K=F*d^2).
Therefore we know that
F(at d=4) = K/4^2, and
F(at d=2) = K/2^2.
Since K is a constant, we can equate K=F*d^2 for each case:
K=F(at d=4)*4^2 = F(at d=2)*2^2
so
F(at d=2) = (F(at d=4)*4^2)/2^2
F(at d=2) = 200N/C * (4^2)/(2^2) = 200N/C *16/4 = 800N/C.
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Most Upvoted Answer
A point charge 50mC is located in the XY plane at the point of positio...
Field intensity is proportional to the inverse of the square of the distance separating the point charges:
F = k*q1*q2/d^2
Since k, q1, and q2 are assumed to be constant, their product can be combined to form a single constant:
K = k*q1*q2, so
F = K/d^2 (and by algebraic manipulation we have K=F*d^2).
Therefore we know that
F(at d=4) = K/4^2, and
F(at d=2) = K/2^2.
Since K is a constant, we can equate K=F*d^2 for each case:
K=F(at d=4)*4^2 = F(at d=2)*2^2
so
F(at d=2) = (F(at d=4)*4^2)/2^2
F(at d=2) = 200N/C * (4^2)/(2^2) = 200N/C *16/4 = 800N/C.
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A point charge 50mC is located in the XY plane at the point of positio...
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