Question

Two charges +5 micro C and +10 micro C are placed 20 cm apart. The net electric feild at the midpoint between the two charges is​

Answer 1

Answer:

the net electric field is 11.25N/C

Answer 2

Answer:

The net electric field at the midpoint between the two charges is $E_{res} = 4.5 \times 10^6J$ and is along 10 micro C to 5 micro C.

Explanation:

Given,

$Q_1 = 5\mu C\\Q_2 = 10\mu C$

separation distance (d) = 20 cm

To find,

Electric field at the midpoint between the charges (E)

Let the electric field due to $Q_1$ be $E_1$, and $Q_2$ be $E_2$

Now,

$E_1 = \frac{kQ_1}{(d/2)^2}$  and   $E_2 = \frac{kQ_2}{(d/2)^2}$

$E_1 = \frac{9\times 10^9\times 5 \mu C}{(10\times 10^{-2})^2}$   and   $E_2 = \frac{9\times 10^9\times 10 \mu C}{(10\times 10^{-2})^2}$

$E_1 = \frac{9\times 10^9\times 5 \times 10^{-6} C}{(10\times 10^{-2})^2}$  and  $E_2 = \frac{9\times 10^9\times 10\times10^{-6} C}{(10\times 10^{-2})^2}$

$E_1 = 4.5 \times 10^6 J$      and      $E_2 = 9\times 10^6 J$

Therefore,

$E_{res} = E_2 - E_1$ (As the electric fields are in opposite direction)

$E_{res} = 9\times 10^6J-4.5 \times 10^6J\\ \implies E_{res} = 4.5 \times 10^6J$

Hence, the net electric field at the midpoint between the two charges is $E_{res} = 4.5 \times 10^6J$ and is along 10 micro C to 5 micro C.

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