Question

Two balls with equal charges are in a vessel with ice at –10⁰C at a distance of 25 cm from each other. On forming water at 0⁰C, the balls are brought nearer to 5 cm for the interaction between them to be same. If the dielectric constant of water at 0⁰C is 80, the dielectric constant of ice at –10⁰C is-

• A. 40
• B. 3.2
• C. 20
• D. 6.4

Answer 1

Answer: (B)

3.2

Answer 2

(–10⁰C ice) (0⁰C water)

$\mathrm { q } _ { 1 } \frac { \mathrm { k } _ { 1 } } { \mathrm { r } _ { 1 } = 25 \mathrm {~cm} } \mathrm { q } _ { 2 }$ $\mathrm { q } _ { 1 } \frac { \mathrm { k } _ { 2 } = 80 } { \mathrm { r } _ { 2 } = 5 \mathrm {~cm} } \mathrm { q } _ { 2 }$

Given F_ice = F_water

$\frac { 1 } { 4 \pi \varepsilon _ { 0 } }$ = $\frac { 1 } { 4 \pi \varepsilon _ { 0 } }$

\ k₁r₁² = k₂r₂²

\ k₁ = k₂ .

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