Question

What is the net charge on a conducting sphere of radius 19cm? given that the electric field 15cm from the centre of the sphere is equal to $3 \times {10^3}N/C$ and is directed inward.
A) $- 7.5 \times {10^{ - 5}}C$
B) $- 7.5 \times {10^{ - 9}}C$
C) $7.5 \times {10^{ - 5}}C$
D) $7.5 \times {10^{ - 9}}C$

Answer 1

Electric field is defined as the force per unit charge. It is the force that is experienced by a unit positive test charge placed at a point. The electric field is inside the conducting sphere and for a conducting sphere the charge goes out to the outer surface but while calculating the net charge on the sphere the electric field is 15cm away and not 19cm.

Complete step by step solution:
The formula for electric field is:
$E = - K\dfrac{Q}{{{r^2}}}$ ;
Here: E = Electric Field;
K = Constant;
Q = Charge;
r = Radius;
Now, rearrange the above equation and solve for the net charge “Q”
Take “Q” from the RHS to the LHS and do the opposite for Electric field “E”.
$\Rightarrow Q = - \dfrac{{E \times {r^2}}}{K}$;
Put in the given value in the above equation and solve:
$Q = - \dfrac{{3 \times {{10}^3} \times {{\left( {15} \right)}^2}}}{{9 \times {{10}^9}}}$;
Do, the necessary calculation:
$\Rightarrow Q = - \dfrac{{{{10}^3} \times 225}}{{3 \times {{10}^9}}}$;
The net charge on a conducting sphere due to an electric field directed inwards is:
$\Rightarrow Q = - 7.5 \times {10^{ - 5}}C$;

Final answer is option A. Therefore, the net charge on a conducting sphere of radius 19 cm is $- 7.5 \times {10^{ - 5}}C$.

Note: Here, we need to apply the formula of electric field which is in relation with the distance, be careful while choosing the distance, we need to calculate the net charge due to electric field and the field is 15cm far. The value of the electric field will be negative as the direction is inwards.