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Byju's Answer

Standard IX

Physics

Electric Potential and Potential Difference

A positive ch...

Question

A positive charge +Q is fixed at a point A. Another positively charged particle of mass m and charge +q is projected from a point B with velocity u as shown in the figure. The point B is at a large distance from A and at a large distance d from the line AC. The initial velocity is parallel to the line AC. The point C is at very large distance from A. Find the minimum distance (in meter) of +q from + Q during the motion. [Take $Q q = 4 π ϵ_{0} m u^{2} d$ ]

d (1 + \sqrt{2})

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d (\sqrt{2} - 1)

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\frac{d}{(1 + \sqrt{2})}

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\frac{d}{(\sqrt{2} - 1)}

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Solution

The correct option is A $d (1 + \sqrt{2})$
Mass $= m$
Charge $= q$
The point $B$ is at a charge distance from $A$ and at a large distance $d$ from the line $A C$
the minimum distance $= d (1 + \sqrt{2})$
$q = (1 + \sqrt{2})$
New distance $= d (1 + \sqrt{2})$ .

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Similar questions

Q. A positive charge +Q is fixed at a point A. Another positively charged particle of mass m and charge +q is projected from a point B with velocity u as shown in the figure. The point B is at large distance from A and at distance ‘d’ from the line AC. The initial velocity is parallel to the line AC. The point C is at very large distance from A. Find the minimum distance (in meter) of +q from +Q during the motion. Take

Q q = 4 π \in_{0} m u^{2} d

and

d = (\sqrt{2} - 1)

meter.

Q. Figure shows a charge

+ Q

clamped at a point in free space. From a large distance another charge particle of charge

- q

and mass of

m

is thrown towards

+ Q

with an impact parameter

d

as shown with speed

v

. How many positions for minimum separation would be attained for the particle.

A charge $q$ of mass $m$ is released with a velocity $1 \times 10^{6} m / s$ from a large distance from a fixed positive charge $Q$ . The closest distance of approach is:

Q. A point charge

(- q)

revolves around a fixed charge

(+ Q)

in an elliptical orbit. The minimum and maximum distance of

- q

from

+ Q

are

r_{1}

and

r_{2}

respectively. The mass of particle

- q

m

and assume no gravitational effects. Find the velocities

v_{1}

and

v_{2}

- q

at positions when it is at

r_{1}

and

r_{2}

distance from

Q

Q. A particle of mass

1 kg

and charge

- 2 μ C

(initially at a very large distance) is projected with initial velocity

200 m/s

towards a positive charged particle of charge

2 μ C

which is fixed. The distance of

1 kg

particle from heavy particle when speed of

1 kg

particle is equal to twice of initial value is

\frac{x}{10^{7}} m

. Find the value of ‘

x

’.

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The correct option is A d(1+√2)Mass =mCharge =qThe point B is at a charge distance from A and at a large distance d from the line ACthe minimum distance =d(1+√2)q=(1+√2)New distance =d(1+√2).

The correct option is A $d (1 + \sqrt{2})$
Mass $= m$
Charge $= q$
The point $B$ is at a charge distance from $A$ and at a large distance $d$ from the line $A C$
the minimum distance $= d (1 + \sqrt{2})$
$q = (1 + \sqrt{2})$
New distance $= d (1 + \sqrt{2})$ .