The magnitude of the charge on the particle is 25 Coulomb.
Solution:
work done = 500 J = W
potential = 20 v =V
charge =q =?
So, by putting the value in the formula, we will get:
500 = q X 20
q = 25 C
Therefore, the magnitude of the charge is 25 coulomb.
The work done is zero because there is no potential difference when a charge moves over an equipotential surface. So the correct option is zero. The SI unit of work is the joule. It is defined as the work done by a force of 1 Newton at a distance of 1 meter. PV work is often measured in units of liter atmosphere.
Where 1L denotes the work done by the system throughout the reversible process. The magnitude of the electric field E produced by a charged particle at a point P is the electric force per unit positive charge exerted on another charged particle at that point. The direction of the electric field is the direction of the force on the positive charge.
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