Question

The magnitude of electric field at point A is 40 NC. The separation of electric field lines at B is twice as that at A. Calculate the force experienced by a proton placed at A and electric field at point B

Answer 1

Answer:

$E_{B} = 20~N/C$

$F_{B} = 6.4~*~10^{-18} N.$

Explanation:

Here, we take into account that Electric field strength at a point is directly proportional to number of electric field lines passing through unit area.

Or, E ∝ $\frac{No.~of~field~lines}{Area}$

When you'll take same areas at point A and point B, you'll find that there're twice the number of electric field lines at point A, compared to B, since the separation between field lines at A is half that that at B.

Hence, Electric field strength at B will be half than that of A.

∴ $\frac{E_{A}}{2} = E_{B}$

⇒$E_{B} = \frac{40}{2}N/C = 20~N/C$

Now, force on a charged particle in an electric field is given as

F = q · E

So, force on a proton at point A will be

F = 1.6 × 10⁻¹⁹ × 40 N

F = 6.4 × 10⁻¹⁸ N.