Question

Two charges +5μC and +10μC are placed 10 cm apart. The net electric field at the mid-point between the two charges is

Answer 1

Answer:

$27 \times 10^{4} N$

Explanation:

Electric field = $\\\tt \dfrac{Force}{charge} = \dfrac{1}{4\pi \epsilon_{\circ}} \times \dfrac{Q}{r^{2}}$

Where Q is the charge and r is the distance between two charges:

Now

At midpoint:

Net electric field =$\\\tt E_{+5\muC} + E_{+10\muC}$

$\\\tt \dfrac{1}{4\pi \epsilon_{\nirc}} \times (\dfrac{10^{-6}}{5\times 10^{-2}} ) (5+10)\\=27 \times 10^{4} N$

Answer 2

Given :-

Two charges +5μC and +10μC are placed 10 cm apart.

To Find :-

The  net electric field at the mid-point between the two charges is

Solution :-

We know that

$\sf Electric \; Field = \dfrac{Force}{Charge}$

$\sf Electric \; Charge = \dfrac{1}{4 \pi e}\times \dfrac{10^{-6}}{5 \times 10^{-2}} \times (5 + 10)$

$\sf Electric \; Charge = \dfrac{1}{4 \pi e}\times \dfrac{10^{-6}}{5 \times 10^{-2}} \times (15)$

$\sf Electric \; Charge = \dfrac{1}{4 \pi e}\times \dfrac{10^{-6}}{10^{-2}} \times (3)$

$\sf Electric \; Charge = 27 \times 10^4$