Question

The diagram shows a uniformly charged hemisphere of radius $R.$ It has volume charge density $\rho .$ If the electric field at a point $2R$ distance above its centre is $E$ then what is the electric field at the point which is $2R$ below its center?

• A. $\frac{\rho R}{6{\epsilon }_0}+E$
• B. $\frac{\rho R}{12{\epsilon }_0}-E$
• C. $-\frac{\rho R}{6{\epsilon }_0}+E$
• D. $\frac{\rho R}{24{\epsilon }_0}+E$

Answer 1

The correct option is B $\frac{\rho R}{12{\epsilon }_0}-E$


Let us complete the sphere. Field at any outside point due to a uniform solid sphere is,

$E=\frac{kq}{r^2}$

$\text{Electric field due to lower part at A}=\text{Electric field due to upper part at B}=E\text{(given)}$

$E_B=\frac{kQ}{(2R{)}^2}-E$

$=\frac{1}{4\pi {\epsilon }_0}\frac{\rho \left(\frac{4}{3}\right)\pi R^3}{4R^2}-E$

$=\frac{\rho R}{12{\epsilon }_0}-E$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.