A parallel plate capacitor was made with two rectangular plates, each with a length of l = 3$l = 3$ cm and breath of b = 1$b = 1$ cm. The distance between the plates is 3mu m$3\mu m$ Out of the following, which are the ways to increase the capacitance by a factor of 10?
A. l = 30$l = 30$ cm, b = 1 cm, d=1mu m$d=1\mu m$
B. l = 3$l = 3$ cm, b=1$b=1$ cm, d=30~mu m$d=30~\mu m$
C. l = 6$l = 6$ cm, b=5$b=5$ cm, d=3~mu m$d=3~\mu m$
D. l = 1$l = 1$ cm, b=1textcm$b=1\text{cm}$, d=10~mu m$d=10~\mu m$
E. l = 5text cm$l = 5\text{ cm}$, b=2$b=2$ cm, d=1mu m$d=1\mu m$
Choose the correct answer from the options given below :
A.C and E only
B.B and D only
C.A only
D.C only
Solution & Explanation
### Related Formula
The capacitance of a parallel plate system is given by :
C = fracepsilon_0Ad = fracepsilon_0lbd$$C = \frac{\epsilon_{0}A}{d} = \frac{\epsilon_{0}lb}{d}$$
where l$l$ is length, b$b$ is breadth, and d$d$ is separation distance.
### Core Logic
Evaluate the initial capacitance base scaling parameter :
C_0 = fracepsilon_0 times 3text cm times 1text cm3mutextm = 1 times epsilon_0text units$$C_{0} = \frac{\epsilon_{0} \times 3\text{ cm} \times 1\text{ cm}}{3\mu\text{m}} = 1 \times \epsilon_{0}\text{ units}$$
We want to increase this initial baseline capacitance value by a factor of 10, meaning our target capacitance is 10epsilon_0$10\epsilon_{0}$.
### Step 1: Audit Options
Let's check the capacitance for options C and E :
* Option C: l=6text cm, b=5text cm, d=3mutextm$l=6\text{ cm}, b=5\text{ cm}, d=3\mu\text{m}$ .
C_C = fracepsilon_0 times 6 times 53 = 10epsilon_0text units (Correct)$$C_{C} = \frac{\epsilon_{0} \times 6 \times 5}{3} = 10\epsilon_{0}\text{ units (Correct)}$$
* Option E: l=5text cm, b=2text cm, d=1mutextm$l=5\text{ cm}, b=2\text{ cm}, d=1\mu\text{m}$ .
C_E = fracepsilon_0 times 5 times 21 = 10epsilon_0text units (Correct)$$C_{E} = \frac{\epsilon_{0} \times 5 \times 2}{1} = 10\epsilon_{0}\text{ units (Correct)}$$
### Pattern Recognition
Capacitance scales matching the geometric factor fracl cdot bd$\frac{l \cdot b}{d}$. Look for options where this ratio scales up to exactly 10 times the initial baseline value.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Keywords:#increase the capacitance by a factor of 10#JEE Main 2025 Morning Q18#Electrostatics JEE Main 2025#Capacitance of a Parallel Plate Capacitor JEE Main 2025
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_2025_24_jan_morning
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