Question

Two charges A and B of 5uC each separated by a distance of 6cm. C is midpoint of the line joining A and B. A charge Q of -5uC is shot perpendicular to the line joining A and B through C with kinetic energy of 0.06J. The charge Q comes to rest at a point D. The distance CD is

Answer 1

Answer on Question #71422 Physics / Electromagnetism

Two charges A and B of $Q = 5\mathrm{uC}$ each separated by a distance of $6\mathrm{cm}$. C is midpoint of the line joining A and B. A charge Q of -5uC is shot perpendicular to the line joining A and B through C with kinetic energy of $KE = 0.06$ J. The charge Q comes to rest at a point D. The distance CD is

Solution:

The potential of electric field at the points and D

$$V_{c} = k \frac{Q}{r_{AC}} + k \frac{Q}{r_{BC}}$$ $$V_{D} = k \frac{Q}{r_{AD}} + k \frac{Q}{r_{BD}}$$

The change of kinetic energy is equal of work done

$$KE = Q(V_{c} - V_{D})$$ $$KE = k Q^{2} \left( \frac{1}{r_{AC}} + \frac{1}{r_{BC}} - \frac{1}{r_{AD}} - \frac{1}{r_{BD}} \right)$$ $$r_{AC} = r_{BC} = \frac{0.06\ \mathrm{m}}{2} \cdot 0.03\ \mathrm{m}$$ $$r_{AD} = r_{BD}$$

Thus

$$0.06 = 9 \times 10^{9} \times (5 \times 10^{-6})^{2} \left( \frac{2}{0.03} - \frac{2}{r_{AD}} \right)$$ $$r_{AD} = 0.03012\ \mathrm{m}$$

Finally

$$CD = \sqrt{ r_{AD}^{2} - r_{AC}^{2} } = 0.0027\ \mathrm{m} = 2.7\ \mathrm{mm}.$$

Answer: $2.7\ \mathrm{mm}$.

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