Question

Two$10\mathrm{cm}$diameter charged rings face each other,$20\mathrm{cm}$apart. Both rings are charged to$+20\mathrm{nC}$. What is the electric field strength at (a) the midpoint between the two rings and (b) the center of the left ring?

Short answer

a.Electric field at midpoint between the two rings is$0\mathrm{N}/\mathrm{C}$.

b. Electric field at the center of the left ring is$4.108\times {10}^3\mathrm{N}/\mathrm{C}$.

Step-by-step solution

Calculation for Electric field at midpoint between the two rings (part a)

(a)

Diameter of both circle is$10\mathrm{cm}$.

Radius is $5\mathrm{cm}$.

Charge on both circle is$+20\mathrm{nC}$.

Distance between both circle is$20\mathrm{cm}$.

Formula used:

Electric field due to a ring of radius$R$and charge$Q$at a distance$r$is given by,

$E=\frac{kQr}{{\left(r^2+R^2\right)}^{3/2}}.......\left(1\right)$

Electric field at the center of both rings will be$0\mathrm{N}/\mathrm{C}$.

Because electric field due to both rings will be equal and in opposite direction.

Figure for given situation

Calculation of electric field at the center of the left ring  (part b)

(b)

Electric field due left circle at its own center is$0\mathrm{N}/\mathrm{C}$.

This can find out by substituting$0$for$\mathrm{r}$in equation$\left(1\right)$.

The electric field generated by the right ring in the middle of the left circle is now in the left direction, and its magnitude may be measured using equation$\left(1\right)$.

$E=\frac{\left(9\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(20\times {10}^{-9}\mathrm{C}\right)\left(20\times {10}^{-2}\mathrm{m}\right)}{{\left((0.2\mathrm{m}{)}^2+(0.05\mathrm{m}{)}^2\right)}^{3/2}}$

$E=4.108\times {10}^3\mathrm{N}/\mathrm{C}$