A point charge +q$+q$ is placed at the origin. A second point charge +9q$+9q$ is placed at (d, 0, 0)$(d, 0, 0)$ in Cartesian coordinate system. The point in between them where the electric field vanishes is:
A.(4d / 3, 0, 0)$(4d / 3, 0, 0)$
B.(d / 4, 0, 0)$(d / 4, 0, 0)$
C.(3d / 4, 0, 0)$(3d / 4, 0, 0)$
D.(d / 3, 0, 0)$(d / 3, 0, 0)$
Solution & Explanation
### Related Formula
E = frack Qr^2$$E = \frac{k Q}{r^2}$$x = fracd1 + sqrtfracq_2q_1$$x = \frac{d}{1 + \sqrt{\frac{q_2}{q_1}}}$$
### Core Logic
Let the null point where the electric field is zero be at (x, 0, 0)$(x, 0, 0)$ where 0 < x < d$0 < x < d$.
At this point, the fields due to both charges are equal in magnitude and opposite in direction:
frack qx^2 = frack (9q)(d - x)^2$$\frac{k q}{x^2} = \frac{k (9q)}{(d - x)^2}$$
Taking the square root on both sides:
frac1x = frac3d - x implies d - x = 3x$$\frac{1}{x} = \frac{3}{d - x} \implies d - x = 3x$$4x = d implies x = fracd4$$4x = d \implies x = \frac{d}{4}$$
Thus, the coordinates of the null point are left(fracd4, 0, 0right)$\left(\frac{d}{4}, 0, 0\right)$.
### Step 1: Final Conclusion
The point where the electric field vanishes is left(fracd4, 0, 0right)$\left(\frac{d}{4}, 0, 0\right)$.
### Pattern Recognition
For two like charges, the zero-field null point always lies along the line joining them and is closer to the smaller charge. Use the standard shortcut: x = fracd1 + sqrtq_2/q_1$x = \frac{d}{1 + \sqrt{q_2/q_1}}$ measured from charge q_1$q_1$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Keywords:#electric field vanishes between point charges#JEE Main 2025 Morning Q7#Electrostatics JEE Main 2025#null point coordinates
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_02_april_morning
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