Question

An uniform electric field E exists along positive x-axis. The work done in moving a charge 0.5 C through
a distance 2 m along a direction making an angle 60° with x-axis is 10 J. Then the magnitude of electric
field is
a) 5 Vm-1
b) 2 Vm-1
c) 5Vm-1
d) 20 V m-1​

Answer 1

Answer:

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Explanation:

5vm-1

Answer 2

The correct answer to the above question is (D), which is 20 Vm⁻¹.

Given: Charge = 0.5C

            Distance = 2m

            Angle = 60°

            Work done is moving a charge is 10J

To find The magnitude of the electric field.

Solution: The electric field at a distance of 2m

• Work done by an electrostatic field in moving a charge from one point to another depends only on the initial and the final points and is independent of the path taken to go from one point to the other

W= Fdcosθ

here, F=qE ( Force in the electromagnetic field is equal to the charge of the particle and electric field in that area.)

W = qEdcosθ

Putting the value into the equation,

W = 0.5×E×2×cos60

E = W/(Cos60×0.5×2)

  =20V/m

Therefore, the electric field is equal to 20 Vm⁻¹.