Question

Two balls with equal charges are in a vessel with ice at $-{10}^{\circ }\text{C}$ at a distance of $25\text{cm}$ from each other. On forming water at $0^{\circ }\text{C},$ the balls are brought nearer to $5\text{cm}$ for the interaction between them to be the same. If the dielectric constant of water at $0^{\circ }\text{C}$ is $80.$ The dielectric constant of ice at $-{10}^{\circ }\text{C}$ is-

• A. $40$
• B. $3.2$
• C. $20$
• D. $6.4$

Answer 1

The correct option is B $3.2$

Given,

Ice is at $-{10}^{\circ }\text{C}$ and water is formed at $0^{\circ }\text{C}$

Force between two identical charges in a medium with dielectric constant $K$ is,

$F_m=\frac{F_{vacuum}}{K}=\frac{q^2}{4\pi {\epsilon }_0\times r^2\times K}$

In water at $0^{\circ }\text{C}$, the force of interaction is,

$F_w=\frac{F_{vacuum}}{K}=\frac{q^2}{4\pi {\epsilon }_0\times (5{)}^2\times 80}$

In ice at $-{10}^{\circ }\text{C}$, the force of interaction is,

$F_{ice}=\frac{F_{vacuum}}{K}=\frac{q^2}{4\pi {\epsilon }_0\times (25{)}^2\times K}$

$\because F_w=F_{ice}$

$\frac{q^2}{4\pi {\epsilon }_0\times (5{)}^2\times 80}=\frac{q^2}{4\pi {\epsilon }_0\times (25{)}^2\times K}$

$\therefore K=80{\left(\frac{5}{25}\right)}^2=3.2$

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