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The electric field intensity at a point P due to point charge q kept at point Q is 24 N C−1 and the electric potential at point P due to same charge is 12 J C−1. The order of magnitude of charge q is
  • a)
    10−6 C
  • b)
    10−7 C
  • c)
    10−10 C
  • d)
    10−9 C
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The electric field intensity at a point Pdue to point charge qkept at ...
Electric field of a point charge.

Electric potential of a point charge,

The distance PQ is r = V/E = 12/24 = 0.5 m
∴ Magnitude of charge
q' = 4πε0Vr

= 10−9 C
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Community Answer
The electric field intensity at a point Pdue to point charge qkept at ...
To find the electric field intensity at point P due to point charge q at point Q, we can use Coulomb's law.

Coulomb's law states that the electric field intensity E at a point in space is given by:

E = k * (q / r^2)

where:
E is the electric field intensity
k is the electrostatic constant (8.99 x 10^9 N m^2/C^2)
q is the charge creating the electric field
r is the distance between the point charge q and the point where the electric field is being measured

Given that the electric field intensity at point P due to point charge q at point Q is 24 N/C, we can set up the equation as follows:

24 = k * (q / r^2)

Now, if we know the value of q and r, we can solve for the electric field intensity E.
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