A point charge of 10^-8$10^{-8}$ C is placed at origin. The work done in moving a point charge 2 mu$\mu$C from point A(4, 4, 2) m to B(2, 2, 1) m is ____ J. left(frac14piepsilon_0=9times10^9text in SI unitsright)$\left(\frac{1}{4\pi\epsilon_{0}}=9\times10^{9}\text{ in SI units}\right)$
A.45 times 10^-6$45 \times 10^{-6}$
B.0$0$
C.30 times 10^-6$30 \times 10^{-6}$
D.15 times 10^-6$15 \times 10^{-6}$
Solution & Explanation
### Related Formula
W_textext = Delta U = U_f - U_i$$W_{\text{ext}} = \Delta U = U_f - U_i$$U = frac14piepsilon_0 fracq_1 q_2r$$U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$$
### Core Logic
Work done by external agent: W_textext = Delta U$W_{\text{ext}} = \Delta U$, where Delta U$\Delta U$ is the change in potential energy.
W_textext = frac14pi epsilon_0 fracq_1 q_2r_f - frac14pi epsilon_0 fracq_1 q_2r_i$$W_{\text{ext}} = \frac{1}{4\pi \epsilon_{0}} \frac{q_{1} q_{2}}{r_{f}} - \frac{1}{4\pi \epsilon_{0}} \frac{q_{1} q_{2}}{r_{i}}$$
Calculate the distances of points A and B from the origin:
r_i = |A| = sqrt4^2 + 4^2 + 2^2 = sqrt16+16+4 = sqrt36 = 6text m$r_i = |A| = \sqrt{4^2 + 4^2 + 2^2} = \sqrt{16+16+4} = \sqrt{36} = 6\text{ m}$r_f = |B| = sqrt2^2 + 2^2 + 1^2 = sqrt4+4+1 = sqrt9 = 3text m$r_f = |B| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3\text{ m}$
### Step 1: Calculate Work Done
W_textext = (9 times 10^9) times (10^-8 times 2 times 10^-6) left[ frac13 - frac16 right]$$W_{\text{ext}} = (9 \times 10^9) \times (10^{-8} \times 2 \times 10^{-6}) \left[ \frac{1}{3} - \frac{1}{6} \right]$$W_textext = 18 times 10^-5 times left(frac2-16right)$$W_{\text{ext}} = 18 \times 10^{-5} \times \left(\frac{2-1}{6}\right)$$W_textext = 18 times 10^-5 times frac16 = 3 times 10^-5text J = 30 times 10^-6text J$$W_{\text{ext}} = 18 \times 10^{-5} \times \frac{1}{6} = 3 \times 10^{-5}\text{ J} = 30 \times 10^{-6}\text{ J}$$
### Pattern Recognition
Electric field is conservative. Work done simply equals change in kqq/r$kqq/r$ from initial to final radial coordinate. No path dependence.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Keywords:#work done in moving a point charge#JEE Main 2026 Morning Q45#Electrostatics JEE Main 2026#Electric Potential Energy JEE Main 2026
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_2026_21_jan_morning
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