Question

Two charges + 5 C and +10 C are placed 20 cm apart from each other. The electric

field at the midpoint of the line joining two charges is….

(a) 4.5 × 106 N/C towards + 5 C (b) 13.5 × 106 N/C towards + 5 C

(c) 4.5 × 106 N/C towards +10 C (d) 13.5 × 106 N/C towards +10 C​

Answer 1

Answer:

$45*10^{11} N/C = 4.5*10^{12}N/C$ Towards the +5 Coulomb charge

Most probably option (a)

Explanation:

Two charges are separated by 20 cm, hence the mid point between charges is placed at 10 cm apart from each charge 10 cm = 0.1 m

The value of electric filed at a point from a charge q at a distance r from the charge is given be the equation below.

$E = \frac{q}{4\pi\epsilon*r^2}$ where $\frac{1}{4\pi\epsilon} = 9*10^9 Nm^2/C^2$

for 5 coulomb charge the value of the electric field at the mid point would be

$\frac{9*10^9*5}{(0.1)^2} = 45*10^{11} N/C=E_1$

Same  for the 10 coulomb charge would be

$\frac{9*10^9*10}{(0.1)^2}= 90*10^{11} N/C = E_2$

10C______________Mid Point________________5C

Now field due to 5C charge $(E_1)$ will act towards left direction (According to the direction.)

& field due to 10C charge $(E_2)$ will act towards right direction (According to the direction.)

Therefore total field would be $E_2-E_1$ at the direction fro left to right

So Total field would be $90*10^{11} - 45*10^{11} = 45*10^{11} N/C$ from 10C to 5 C charge

Answer 2

Given:

Two charges 5 × 10^(-8)C and 10 ×10^(-8)C are placed 20 cm apart from each other.

To find:

Net field intensity at midpoint ?

Calculation:

At midpoint, the field intensity due to individual charges will be opposite to one another. Since the field due to 10 × 10^(-8)C will be stronger, the net field will be towards +5×10^(-8)C.

$E = \bigg \{ \dfrac{k(10)}{ {(0.01)}^{2} } - \dfrac{k(5)}{ {(0.01)}^{2} } \bigg \}\times {10}^{ - 8}$

$\implies E = \dfrac{k(5)}{ {(0.01)}^{2} } \times {10}^{ - 8}$

$\implies E = \dfrac{9 \times {10}^{9} \times 5}{ {10}^{ - 4} } \times {10}^{ - 8}$

$\implies E = 45 \times {10}^{5}$

$\implies E = 4.5 \times {10}^{6} \: N/C$

So, field intensity is 4.5 × 10⁶ towards +5 × 10^(-8) C charge.