Question

a force of 5N is acting on the charge 6uC at any point. Determine the electric field intensity at that point?​

Answer 1

The electric field intensity at that point is 8.34×10⁵ N/C.

Given:

Force, F =5 N

Charge, q = 6 μC

To Find:

The electric field intensity at that point.

Solution:

We are required to find the electric field intensity at that point.

The electric field intensity at a point is the force experienced by a unit of positive charge placed at that point.

The electric field intensity at that point is given as

Electric field, E = F/q ------(1)

Force, F =5 N

Charge, q = 6 μC

Substitute the value of Force(F) and charge(q) in equation(1)

Electric field, E = 5 / 6×10⁻⁶

E = 0.833333×10⁶

E = 8.34×10⁵ N/C

Therefore, The electric field intensity at that point is 8.34×10⁵ N/C.

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