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प्रश्न
Two charges 5 × 10−8 C and −3 × 10−8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
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उत्तर
Given: q1 = 5 × 10−8 C
q2 = −3 × 10−8 C
r = 16 cm
Let C be the point on the line joining the two charges, where the electric potential is zero, then
VC = 0
VCA + VCB = 0
VCA = −VCB
`1/(4piε_0) q_A/r_(CA) = -1/(4piε_0) q_B/r_(CB)`
`(5 xx 10^-8 C)/(x xx 10^-2 m) = (-(-3 xx 10^-8 C))/([(16 - x) xx 10^-2 m])`
`5/x = 3/(16 -x)`
5(16 − x) = 3x
80 − 5x = 3x
8x = 80 m
∴ `x = 80/8`
= 10 cm
So, the electric potential is zero at a distance of 10 cm from the charge 5 × 10−8 C on the line joining the two charges, between them.
If point C is not between the two charges, then
VCA + VCB = 0
VCA = −VCB
`1/(4piε_0) q_A/r_(CA) = -1/(4piε_0) q_B/r_(CB)`
`(5 xx 10^-8 C)/([(16 + x) xx 10^-2 m]) = (-(-3 xx 10^-8 C))/(x xx 10^-2 m)`
`5/(16 + x) = 3/x`
5x = 48 + 3x
2x = 48 m
= 24 cm
So, the electric potential is also zero at a distance of 24 cm from the charge −3 × 10−8 C, i.e., at a distance of (24 + 16) = 40 cm from the charge 5 × 10−8 C on the side of the charge −3 × 10−8 C.
संबंधित प्रश्न
An electric dipole is held in a uniform electric field.
(i) Show that the net force acting on it is zero.
(ii) The dipole is aligned parallel to the field. Find the work done in rotating it through the angle of 180°.
Consider the situation shown in figure (31-E29). The width of each plate is b. The capacitor plates are rigidly clamped in the laboratory and connected to a battery of emf ε. All surfaces are frictionless. Calculate the value of M for which the dielectric slab will stay in equilibrium.
Derive an expression for the electric potential at any point along the axial line of an electric dipole.
Which of the following is correct for energy of a dipole in a uniform electric field?
The electric potential at a point on the equatorial line of an electric dipole is ______.
A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. The potential at a distance of 2 cm from the centre of the sphere is ______.
An electric field of 1000 V/m is applied to an electric dipole at angle of 45°. The value of electric dipole moment is 10-29 C.m. What is the potential energy of the electric dipole?
A point P lies at a distance x from the midpoint of an electric dipole on its axis. The electric potential at point P is proportional to ______.
Assertion (A): The potential (V) at any axial point, at 2 m distance (r) from the centre of the dipole of dipole moment vector `vec P` of magnitude, 4 × 10−6 cm, is ±9 × 103 V.
(Take `1/(4 pi epsilon_0)` = 9 × 109 SI units)
Reason (R): V = `+- (2 vec P)/(4 pi epsilon_0 r^2)`, where r is the distance of any axial point, situated at 2 m from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below:
Assertion A: The potential (V) at any axial point, at 2 m distance (r) from the centre of the dipole of dipole moment vector of magnitude, 4 × 10−6 Cm, is ± 9 × 103 V. (Take = 9 × 109 SI units)
Reason R: where r is the distance of any axial point, situated at 2 m from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below:
