A small uncharged conducting sphere is placed in contact with an identical sphere but having 4 times 10^-8$4 \times 10^{-8}$ C charge and then removed to a distance such that the force of repulsion between them is 9 times 10^-3$9 \times 10^{-3}$ N. The distance between them is (Take frac14pivarepsilon_0$\frac{1}{4\pi\varepsilon_{0}}$ as 9 times 10^9$9 \times 10^{9}$ in SI units)

### Related Formula E = frackqx^2$$E = \frac{kq}{x^2}$$ ### Core Logic

For the net electric field to be zero at point P situated at distance x$x$ from charge q$q$, the electric fields produced by both charges must be equal in magnitude and opposite in direction. Let the charges be placed at ends of a line. Point P is between them since both charges are of the same sign. (vecE_textnet)_P = 0$$(\vec{E}_{\text{net}})_P = 0$$ frackqx^2 = frack(3q)(r-x)^2$$\frac{kq}{x^2} = \frac{k(3q)}{(r-x)^2}$$ ### Step 2: Solving for x Taking square roots on both sides: frac1x = fracsqrt3r-x$$\frac{1}{x} = \frac{\sqrt{3}}{r-x}$$ r - x = sqrt3x$$r - x = \sqrt{3}x$$ r = x(sqrt3 + 1)$$r = x(\sqrt{3} + 1)$$ x = fracrsqrt3 + 1$$x = \frac{r}{\sqrt{3} + 1}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula C = fracvarepsilon_0 Ad$$C = \frac{\varepsilon_0 A}{d}$$ C' = fracvarepsilon_0 Ad - t + fractK$$C' = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$$ Q = CV$Q = CV$ ### Core Logic Initially, the charge stored without the dielectric is: Q_i = fracA varepsilon_0d V$$Q_i = \frac{A \varepsilon_0}{d} V$$ After introducing a dielectric of thickness t$t$, the new capacitance C'$C'$ leads to a new charge Q_f$Q_f$: Q_f = fracA varepsilon_0 Vd - t + fractK$$Q_f = \frac{A \varepsilon_0 V}{d - t + \frac{t}{K}}$$ ### Step 2: Charge Relationship Given that the capacitor draws 25\%$25\%$ more charge: Q_f = 1.25 Q_i = frac54 Q_i$$Q_f = 1.25 Q_i = \frac{5}{4} Q_i$$ Equating the expressions: fracA varepsilon_0 Vd - t + fractK = 1.25 left( fracA varepsilon_0 Vd right)$$\frac{A \varepsilon_0 V}{d - t + \frac{t}{K}} = 1.25 \left( \frac{A \varepsilon_0 V}{d} \right)$$ frac15 - 2 + frac2K = frac1.255$$\frac{1}{5 - 2 + \frac{2}{K}} = \frac{1.25}{5}$$ frac13 + frac2K = frac1.255 = frac14$$\frac{1}{3 + \frac{2}{K}} = \frac{1.25}{5} = \frac{1}{4}$$ 3 + frac2K = 4$$3 + \frac{2}{K} = 4$$ frac2K = 1 Rightarrow K = 2$$\frac{2}{K} = 1 \Rightarrow K = 2$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics