Ask & Answer Question

Electrostatic Potential Physics Electrostatic Potential and Capacitance Physics Class 12th

Two point charge Q each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulomb’s force is:

Option 1 -

x = d

Option 2 -

$x = \frac{d}{2}$

Option 3 -

$x = \frac{d}{\sqrt[]{2}}$

Option 4 -

$x = \frac{d}{2 \sqrt[]{2}}$

4 Views | Posted 2 months ago

Asked by Shiksha User

2 Answers

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

According to question, we can write

$F_{C A} = F_{C B} = \frac{K q Q}{x^{2} + {(\frac{d}{2})}^{2}} \Rightarrow F = 2 F_{C A} c o s θ = \frac{2 K q Q x}{{[x^{2} + {(\frac{d}{2})}^{2}]}^{\frac{3}{2}}}$

For maxima of force

$\frac{d F}{d x} = 0, s o$

$x = \frac{d}{2 \sqrt[]{2}}$

0 Upvotes 0 Downvotes

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

According to question, we can write

$F_{C A} = F_{C B} = \frac{K q Q}{x^{2} + {(\frac{d}{2})}^{2}} ? F = 2 F_{C A} c o s ? = \frac{2 K q Q x}{{[x^{2} + {(\frac{d}{2})}^{2}]}^{\frac{3}{2}}}$

For maxima of force

$\frac{d F}{d x} = 0, ? ? s o$

$x = \frac{d}{2 \sqrt[]{2}}$

0 Upvotes 0 Downvotes

Similar Questions for you

A particle of charge q and mass m is subjected to an electric field E = E₀(1 − ax²) in the x-direction, where a and E₀ are constants. Initially the particle was at rest at x = 0. Other then the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is:

Vishal Baghel

E = E? (1 − ax²)

F = qE?
acceleration = F/m = (qE? /m) (1 - ax²) = v (dv/dx)
(qE? /m) ∫? (1 - ax²)dx = ∫? vdv; (qE? /M) (x - ax³/3) = 0
x (1 - ax²/3) = 0; x = 0 & x = √ (3/a)

A capacitor of capacitance C = 1µF is suddenly connected to a battery of 100 volt through a resistance R = 100Ω. The time taken for the capacitor to be charged to get 50 V is: [Take ln 2 = 0.69]

Vishal Baghel

V = E (1 - e? /τ)
Where τ = RC = 100 × 10? = 10? sec
50V = 100V (1 - e? /10? )
1/2 = 1 - e? ¹?
1/2 = e? ¹?
-ln2 = -10? t
t = ln2/10?
t = 0.693 × 10? sec

A two point charges $4 q$ and $- q$ are fixed on the $x$ -axis at $x = - \frac{d}{2}$ and $x = \frac{d}{2}$ , respectively. If a third point charge ' $q$ ' is taken from the origin to $x = d$ along the semicircle as shown in the figure, the energy of the charge will:

alok kumar singh

$U_{initial} = \frac{k (4 q) (q)}{(d / 2)} + \frac{k (q) (- q)}{(d / 2)}$

$\begin{array}{r} \Rightarrow \frac{6 k q^{2}}{d} \\ \Rightarrow U_{final} = \frac{4 (4 q) (q)}{(\frac{3 d}{2})} + \frac{k (q) (- q)}{(d / 2)} \\ \Rightarrow \frac{2}{3} \frac{k q^{2}}{d} \\ \Rightarrow Δ U = (\frac{2}{3} - 6) \frac{k q^{2}}{d} \Rightarrow \frac{- 16}{3} \frac{{k q}^{2}}{d} \end{array}$

A air bubble of radius 1 cm in water has an upward acceleration 9.8 cm^-2 . The density of water is 1 gmcm^-3 and water offers negligible drag force on the bubble. The mass of the bubble is: (g=980 cm/s²)

alok kumar singh

$ρ v g - m g = m a$

$\Rightarrow \frac{ρ v g}{m} = g + a$

$\Rightarrow m = \frac{ρ v g}{g + a}$

$\Rightarrow \frac{10^{3} (\frac{4}{3} π \times 10^{- 6}) (9.8)}{9.898}$

$\Rightarrow 4.15 g m$

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Two point charge Q each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulomb’s force is:

2 Answers

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question