Question

Two rings of radius $\mathbf{5}\mathbf{}\mathbf{cm}$are $\mathbf{20}\mathbf{}\mathbf{cm}$apart and concentric with a common horizontal axis. The ring on the left carries a uniformly distributed charge of $\mathbf{+}\mathbf{35}\mathbf{nC}$, and the ring on the right carries a uniformly distributed charge of $\mathbf{-}\mathbf{35}\mathbf{nC}$. (a) What are the magnitude and direction of the electric field on the axis, halfway between the two rings? (b) If a charge of$\mathbf{-}\mathbf{5}\mathbf{nC}$were placed midway between the rings, what would be the magnitude and direction of the force exerted on this charge by the rings? (c) What are the magnitude and direction of the electric field midway between the rings if both rings carry a charge of $\mathbf{+}\mathbf{35}\mathbf{nC}$?

Short answer

(a) The magnitude of the electric field on the axis, halfway between the two rings, isin the right direction.

(b) The magnitude of the force exerted on this charge by the rings is,$-2.25\times {10}^{-4}\mathrm{N}$ and it is towards the left.

(c) Thus, the magnitude of the electric field if both rings carry the same charge is 0 and that is why no direction is there

Step-by-step solution

identification of the given data

The given data is listed below as,

• The radius of the rings is,$\mathrm{R}=5\mathrm{cm}=5\mathrm{cm}\times \frac{1\mathrm{cm}}{100\mathrm{cm}}=0.05\mathrm{m}$
• The distance between the rings is, $\mathrm{r}=20\mathrm{cm}=20\mathrm{cm}\times \frac{1\mathrm{cm}}{100\mathrm{cm}}=0.2\mathrm{m}$
• The uniformly distributed charge on the left ring is,${\mathrm{q}}_1=+35\mathrm{nC}$
• The uniformly distributed charge on the right ring is,${\mathrm{q}}_2=-35\mathrm{nC}$

Significance of the electric field on a ring and force due to electric field

The fields of a point charge of the element having infinitesimal charge can be used to obtain the electric field on the ring by superimposing the charge fields.

The force due to the electric field can be determined by taking the product of the charge on the object and the electric field.

 Step 3: (a) Determination of the magnitude and direction of the electric field on the axis

The expression of the magnitude of the electric field at the midway point is expressed as follows,

$E=\frac{1}{4\pi {\epsilon }_0}\frac{\left(q_1-q_2\right)r}{{\left(R^2+r^2\right)}^{{\displaystyle \frac{3}{2}}}}$

Here,$\frac{1}{4\pi {\epsilon }_0}$is the constant of the electric field, and its value is , is the charge on the left ring, is the charge on the right ring, is the radius of the ring, and is the center distance between the rings.

Substitute all the values in the above expression.

$$\begin{gathered} E=9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\times \frac{\left(35\mathrm{nC}-\left(-35\mathrm{nC}\right)\right)\times \left(0.2\mathrm{cm}/2\right)}{{\left({\left(0.05\mathrm{m}\right)}^2+{\left(0.02\mathrm{cm}/2\right)}^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ =9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\times \frac{\left(35\times {10}^{-9}\mathrm{C}-\left(-35\times {10}^{-9}\mathrm{C}\right)\right)\times 0.1\mathrm{m}}{{\left({\left(0.05\mathrm{m}\right)}^2+{\left(0.1\mathrm{m}\right)}^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ =9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\times \frac{\left(\left(7\times {10}^{-8}\mathrm{C}\right)\right)\times 0.1\mathrm{m}}{1.39\times {10}^{-3}\mathrm{m}} \\ =9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\times \frac{7\times {10}^{-9}\mathrm{C}.\mathrm{m}}{1.39\times {10}^{-3}\mathrm{m}} \\ =9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\times 5.03\times {10}^{-6}\mathrm{C}/{\mathrm{m}}^2 \\ =4.5\times {10}^4\mathrm{N}/\mathrm{C} \end{gathered}$$

As the electric field is positive, then the electric field is in the right direction.

Thus, the magnitude of the electric field on the axis, halfway between the two rings, is in the right direction.

(b) Determination of the magnitude and direction of the electric field exerted by a particular charge

The expression for the force exerted is expressed as,

$F=QE$

Here,$Q$is the charge on the object and,$E$is the electric field.

Substitute all the values in the above expression.

localid="1656931069322" $$\begin{gathered} F=\left(-5\mathrm{nC}\times \frac{1\mathrm{C}}{1\times {10}^9\mathrm{nC}}\right)\times \left(4.5\times {10}^{-4}\mathrm{N}/\mathrm{C}\right) \\ =\left(-5\times {10}^{-9}\mathrm{C}\right)\times \left(4.5\times {10}^4\mathrm{N}/C\right) \\ =-2.25\times {10}^{-4}\mathrm{N} \end{gathered}$$

As the force is negative, then the force’s direction is in the left direction.

Thus, the magnitude of the force exerted on this charge by the rings is towards the left.

(c) Determination of the magnitude and the direction of the electric field if both rings carry the same charge

The expression for the magnitude of the electric field at the midway point is expressed as,

$E=\frac{1}{4\pi {\epsilon }_0}\frac{\left(q_{1-}{q}_2\right)r}{{\left(R^{2+}{r}^2\right)}^{{\displaystyle \frac{3}{2}}}}$

If both the rings carry the same charge, then according to the equation of the electric field, the magnitude of the electric field is zero.

Thus, the magnitude of the electric field if both rings carry the same charge is 0 and that is why no direction is there.