Question

Two semicircular wires of radius $20$cm and $10$cm have a common centre at the origin $O$as shown in the figure. Assume that both the wires are uniformly charged and have an equal charge $0.70nC$ each. The magnitude of electric field ( in $V{{m}^{-1}}$) at the common centre of curvature O of the system is

A.$100$ v/m
B.$301$ v/m
C.$401$ v/m
D.$501$ v/m

Answer 1

Hint:In this question we have to find out the magnitude of electric field strength at the common centre of curvature O . The charge is distributed over the semicircle. We can not use coulomb's rule here. Hence we have to use the formula of the electric field at the centre of the semicircle.





Formula used:
The electric field strength is related to the total charge of the semicircle by the following formula:
$E=\dfrac{2kq}{\pi {{R}^{2}}}$
Here E = Electric field strength
k = constant with a value $8.99\times {{10}^{9}}$ $N{{m}^{2}}/s$
q = total charge and the distance of the point from the charged surface area




Complete answer:
Let the radius of two semicircles be ${{R}_{1}}$ and r. Both circles share a common centre O and ${{E}_{1}},{{E}_{2}}$ are the opposite electric field vectors along the x-axis. Thus they cancel each other, and only the electric field vector along the Y axis will contribute to the total magnitude of the electric field, ${{E}_{net}}$.

 Both the wires are uniformly charged and have an equal charge of $0.70$ nC each.
Given ${{R}_{1}}$$=20$ cm $=0.2m$ and r$=10$ cm$=0.1m$
q$=0.7$ nC $=0.7\times {{10}^{-9}}$ C; since $1$ nC$=1\times {{10}^{-9}}C$
For the semicircle with radius ${{R}_{1}}$, the electric field strength, ${{E}_{{{R}_{1}}}}=\dfrac{2kq}{\pi R_{1}^{2}}$
And for the semicircle with radius r, the electric field strength,${{E}_{r}}=\dfrac{2kq}{\pi {{r}^{2}}}$
Therefore ${{E}_{net}}={{E}_{r}}-{{E}_{{{R}_{1}}}}$
${{E}_{net}}=\dfrac{2kq}{\pi {{r}^{2}}}-\dfrac{2kq}{\pi R_{1}^{2}}$
${{E}_{net}}=\dfrac{2kq}{\pi }\left( \dfrac{1}{{{r}^{2}}}-\dfrac{1}{{{R}_{1}}^{2}} \right)$
Putting the above values we get,
${{E}_{net}}=\dfrac{2\times 8.99\times {{10}^{9}}\times 0.7\times {{10}^{-9}}}{3.14}\left( \dfrac{1}{{{(0.1)}^{2}}}-\dfrac{1}{{{(0.2)}^{2}}} \right)$
${{E}_{net}}=\dfrac{2\times 8.99\times {{10}^{9}}\times 0.7\times {{10}^{-9}}}{3.14}\times 75$
${{E}_{net}}=300.96$ $V{{m}^{-1}}$ $\approx 301$ v/m
Therefore, the magnitude of the electric field at the common centre of curvature O of the system is $301$ $V{{m}^{-1}}$ .
Thus option (B) is correct.





Note:Here it is important to remember the condition ${{E}_{Y}}={{E}_{net}}$ I.e, only the y-axis will contribute to the total field and the component x will come out to be zero. This happens due to symmetry and electric fields being vectors, and all the x values are equal in magnitude but opposite in direction.