Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle theta$\theta$ with each other. When suspended in water the angle remains the same. If density of the material of the sphere is 1.5mathrm~g/cc$1.5\mathrm{~g/cc}$, the dielectric constant of water will be (Take density of water = 1mathrm~g/cc$= 1\mathrm{~g/cc}$):
Numerical Answer Type:
Enter a numerical valueAnswer: 3 to 3+4 marks
Solution & Explanation
### Related Formula
Equilibrium condition for electrostatic suspension:
tanleft(fractheta2right) = fracF_emg = fracq^24pivarepsilon_0 r^2 mg$$\tan\left(\frac{\theta}{2}\right) = \frac{F_e}{mg} = \frac{q^2}{4\pi\varepsilon_0 r^2 mg}$$
In a liquid medium with buoyant force mitigation:
tanleft(fractheta2right) = fracF_e'mg_texteff = fracq^24pivarepsilon_0 varepsilon_r r^2 mg left(1 - fracrho_textliquidrho_textsolidright)$$\tan\left(\frac{\theta}{2}\right) = \frac{F_e'}{mg_{\text{eff}}} = \frac{q^2}{4\pi\varepsilon_0 \varepsilon_r r^2 mg \left(1 - \frac{\rho_{\text{liquid}}}{\rho_{\text{solid}}}\right)}$$
### Core Logic
Since the angle theta$\theta$ stays exactly the same in both scenarios, we can equate the two balance ratios:
fracF_emg = fracF_e'mg_texteff implies 1 = varepsilon_r left(1 - fracrho_wrho_sright)$$\frac{F_e}{mg} = \frac{F_e'}{mg_{\text{eff}}} \implies 1 = \varepsilon_r \left(1 - \frac{\rho_w}{\rho_s}\right)$$
### Step 1: Substitute Densities
Given data: rho_s = 1.5mathrm~g/cc$\rho_s = 1.5\mathrm{~g/cc}$, rho_w = 1.0mathrm~g/cc$\rho_w = 1.0\mathrm{~g/cc}$.
1 = varepsilon_r left(1 - frac11.5right) = varepsilon_r left(1 - frac23right) = varepsilon_r left(frac13right)$$1 = \varepsilon_r \left(1 - \frac{1}{1.5}\right) = \varepsilon_r \left(1 - \frac{2}{3}\right) = \varepsilon_r \left(\frac{1}{3}\right)$$varepsilon_r = 3$$\varepsilon_r = 3$$
### Pattern Recognition
Shortcut formula for invariant angle setups:
varepsilon_r = fracrho_textsolidrho_textsolid - rho_textliquid = frac1.51.5 - 1 = frac1.50.5 = 3$$\varepsilon_r = \frac{\rho_{\text{solid}}}{\rho_{\text{solid}} - \rho_{\text{liquid}}} = \frac{1.5}{1.5 - 1} = \frac{1.5}{0.5} = 3$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Class 11 Physics: Mechanical Properties of Fluids
Free body diagrams for electrostatic suspension in fluid medium Q54
Keywords:#Dielectric constant calculation fluid immersion#JEE Main 2024 Morning Q54#Coulomb force buoyancy string balance#Electrostatic liquid permittivity factor
More Electrostatics Previous-Year Questions — Page 9
Q42jee_main_2024_31_jan_morningElectric Field Zero Point
Two charges q$q$ and 3q$3q$ are separated by a distance 'r$r$' in air. At a distance x$x$ from charge q$q$, the resultant electric field is zero. The value of x$x$ is :
A.frac(1 + sqrt3)r$\frac{(1 + \sqrt{3})}{r}$
B.fracr3(1 + sqrt3)$\frac{r}{3(1 + \sqrt{3})}$
C.fracr(1 + sqrt3)$\frac{r}{(1 + \sqrt{3})}$
D.r(1 + sqrt3)$r(1 + \sqrt{3})$
Solution
### Related Formula
E = frackqx^2$$E = \frac{kq}{x^2}$$
### Core Logic
Electric Field Zero Point diagram for Q42 - JEE Main 2024 Morning
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
Q52jee_main_2024_31_jan_morningCapacitance With Dielectric
A parallel plate capacitor with plate separation5 mathrm~mm$5 \mathrm{~mm}$ is charged up by a battery. It is found that on introducing a dielectric sheet of thickness 2 mathrm~mm$2 \mathrm{~mm}$, while keeping the battery connections intact, the capacitor draws 25 \%$25 \%$ more charge from the battery than before. The dielectric constant of the sheet is _____.
Numerical Answer.Answer: 2 to 2
Solution
### 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
More Electrostatics Questions — jee_main_2024_01_february_morning
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