The potential energy of a particle in a force field is: `U = (A)/(r^(2)) - (B)/(r )`,. Where `A` and `B` are positive constants and `r` is the distance of particle from the centre of the field. For stable equilibrium the distance of the particle isA. `B//A`B. `B//2A`C. `2A//B`D. `A//B`

The potential energy of a particle in a force field is: `U = (A)/(r^(2

1.	The potential energy of a particle in a force field is: `U = (A)/(r^(2)) - (B)/(r )`,. Where `A` and `B` are positive constants and `r` is the distance of particle from the centre of the field. For stable equilibrium the distance of the particle isA. `B//A`B. `B//2A`C. `2A//B`D. `A//B`
Answer» Correct Answer - C Force experienced by the particle in field, `F = - ((dU)/(dr)) = - ((-2A)/(r^(3))+(B)/(r^(2)))` At equilibrium `F = 0 F = - ((-2A)/(r^(3))+(B)/(r^(2))) = 0` `rArr r = (2A)/(B)`

Discussion

No Comment Found

Related InterviewSolutions

If the potential of a capacitor having capacity of `6 muF` is increased from 10 V to 20 V,then increase in its energy will beA. `12 xx 10^(-6) J`B. `9 xx 10^(-4) J`C. `4 xx 10^(-6) J`D. `4 xx 10^(-9) J`
The electric potential V at any point x,y,z (all in metre) in space is given by `V=4x^2` volt. The electric field at the point `(1m, 0, 2m)` is ……………`V/m`.A. 8 along negative X-axisB. 8 along positive X-axisC. 16 along negative X-axisD. 16 along positive Z-axis
Electric potential is given by `V = 6x - 8xy^(2) - 8y + 6yz - 4z^(2)` Then electric force acting on `2C` point charge placed on origin will beA. 2 NB. 6 NC. 8 ND. 20 N
Variation of electrostatic potential along the x - direction is shown in (Fig. 3.142). The correct statement about electric field is .A. x - component at point `B` is maximumB. x - component at point `A` is toward positive x - axisC. x - component at point `C` is along negative x - axisD. x - component at point `C` is along positive x - axis
The potentaial function of an electrostatic field is given by `V = 2 x^2`. Determine the electric field strength at the point `(2 m, 0, 3 m)`.A. `vec E = 4 hat i(NC^-1)`B. `vec E = -4 hat i(NC^-1)`C. `vec E = 8 hat i(NC^-1)`D. `vec E = -8 hat i(NC^-1)`
A ball of mass `1g` and charge `10^(-8) C` moves from a point `A`. Where potential is `600` volt to the point `B` where potential is zero. Velocity of the ball at the point `B` is `20 cm//s`. The velocity of the ball at the point `A` will beA. `22.8 cm//s`B. `228cm//s`C. `16.8 cm//s`D. `168 cm//s`
Find the ratio of electric work done in bringing a charge `q` from `A` to `B(W_(AB))` and that from `B` to `C(W_(BC))` in a sphere of charge `Q` distributed uniformly throughout its volume. A. `1`B. `1.5`C. `0.75`D. None of these
The work done in bringing a `20` coulomb charge from point `A` to point `B` for disatnce `0.2 m` is `2 J`. The potential difference between the two points will be (in volt)A. `0.2`B. `8`C. `0.1`D. `0.4`
Assertion: If the distance between parallel plates of a capacitor is halved and dielectric constant is made three times, then the capacitor becomes `6` times. Reason: Capacity of the capacitor does not depend upon the nature of the meterial.A. If both assertion and reason are true and reson is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If the assertion is true but the reason are false.D. If both the assertion and reason are false.
A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is ismilarly charged to a potential difference 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the poistive terminal of one is connected to the negative terminal of the other. The final energy of the configuration isA. ZeroB. `(23CV^(2))/(6)`C. `(3CV^(2))/(2)`D. `(9CV^(2))/(2)`

The potential energy of a particle in a force field is: `U = (A)/(r^(2)) - (B)/(r )`,. Where `A` and `B` are positive constants and `r` is the distance of particle from the centre of the field. For stable equilibrium the distance of the particle isA. `B//A`B. `B//2A`C. `2A//B`D. `A//B`

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment