Question

There are two charges 5 μC and 2 μC, placed at some finite distance. The ratio of electrostatic field strengths at midpoint on the line joining between them will be

Answer 1

Answer:

5:2

Explanation:

I think the answer will be 5:2 Why?

Because,

as you are talking about the midpoint this means that the distance from that point to both the charges is equal. So,

field strengths would respectively be (constant*5)/r²

and (constant*2)/r² (since, r is equal)

so the ratio =5:2

Answer 2

Answer:

The ratio of electrostatic field strengths is 5:2.

Explanation:

The electrostatic field for a point charge is given as,

$E=\frac{kQ}{d^2}$       (1)

Where,

E=electrostatic field

k=coulombs constant

Q=charge on the particle

d=distance at which the electrostatic field is to be found

From the question we have,

Q₁=5μC=5 ×10⁻⁶C

Q₂=2μC=2×10⁻⁶C

For the charge, the Q₁ E field will be,

$E_1=\frac{k\times 5\times 10^-6}{(\frac{r}{2} )^2}$       (2)       (∵d=r/2 )

For the charge, the Q₂ E field will be,

$E_2=\frac{k\times 2\times 10^-6}{(\frac{r}{2} )^2}$        (3)

By taking the ratio of equations (2) and (3) we get;

$\frac{E_1}{E_2} =\frac{\frac{k\times 5\times 10^-6}{(\frac{r}{2} )^2}}{\frac{k\times 2\times 10^-6}{(\frac{r}{2} )^2}}$

$\frac{E_1}{E_2} =\frac{5\times 10^-^6}{2\times 10^-^6} =\frac{5}{2}$

Hence, the ratio of electrostatic field strengths at the midpoint on the line joining between the charges is 5:2.