Question

Two charges Q₁and Q₂ are distance d apart. Two dielectrics of thickness t₁ and t₂ and dielectric constant k₁ and k₂ are introduced as shown. Find the force between the charges –

• A. $\frac{Q_{1}Q_{2}}{4\pi\varepsilon_{0}\lbrack d–(t_{1} + t_{2}) + k_{1}t_{1} + k_{2}t_{2}\rbrack^{2}}$
• B. Zero
• C. $\frac{Q_{1}Q_{2}}{4\pi\varepsilon_{0}\lbrack d + \sqrt{k_{1}}t_{1} + \sqrt{k_{2}}t_{2}\rbrack^{2}}$
• D. $\frac{Q_{1}Q_{2}}{4\pi\varepsilon_{0}\lbrack\sqrt{k_{1}}t_{1} + \sqrt{k_{2}}t_{2} + d–(t_{1} + t_{2})\rbrack^{2}}$

Answer 1

Answer: (D)

$\frac{Q_{1}Q_{2}}{4\pi\varepsilon_{0}\lbrack\sqrt{k_{1}}t_{1} + \sqrt{k_{2}}t_{2} + d–(t_{1} + t_{2})\rbrack^{2}}$

Answer 2

Effective distance in vacuum

= $\sqrt { \mathrm { k } _ { 1 } }$ t₁ + $\sqrt { \mathrm { k } _ { 2 } }$ t₂ – (t₁ + t₂).