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If an electron moves from rest from a point at which potential is 50 volt to another point at which potential is 70 volt, then its kinetic energy in the final state will be .....
$ K.E = q_o (V_A - V_B ) = 1.6 \times 10^{-6} (70- 50) = 3.2 \times 10^{-18 } J $
Two electric charges 12 c and –6 c are placed 20cm apart in air. There will be a point P on the line joining these charges and outside the region between them, at which the electric potential is zero. The distance of P from –6 c charge is
4 Points charges each +q is placed on the circumference of a circle of diameter 2d in such a way that they form a square. The potential at the centre is ......
Calculaate as MCQ 66
Three identical charges each of 2 c are placed at the vertices of a triangle ABC as shown in the figure. If AB + AC = 12 cm and $ AB . AC = 32cm^2$, the potential energy of the charge at A is .....
$AB +AC = 12 cm ...(1)
A ball of mass 1 gm and charge $10^{–8} c $ moves from a point A, where the potential is 600 volt to the point B where the potential is zero. Velocity of the ball of the point B is 20cm/s. The velocity of the ball at the point A will be .....
Use the equation $ 1/ 2 m ( V_1^2 - V_2^2 ) = QV $
A thin spherical conducting shell of radius R has a charge q. Another charge Q is placed at the centre of the shell. The electrostatic potential at a point p a distance R/2 from the centre of the shell is .....
When charge q is released in uniform electric field E then its acceleration $ a = { qE \over m } $ (is constant) so it’s motion will be uniformly accelerated motion and it’s Velocity after time is given by $ V = at = { qE \over m } t = K = { 1/2 } mv^2 = {1/2 } ( {qq\over m} t )^2 = { q^2 E^2 t^2 \over 2m }$
Two point charges –q and +q are located at points (o, o, –a) and (o, o, a) respectively. The potential at a point (o, o, z) where z > a is
The potential at a point (0, 0, z) due to a dipole formed by charges -q at (0, 0, -a) and +q at (0, 0, a) can be calculated using the formula for the potential due to a dipole. The correct formula for the potential at a distance z from the dipole center along the axis is V = (1/4πε₀) * (2qa / (z² - a²)). Therefore, the correct answer is $\frac{2qa}{4\pi\varepsilon_0(z^2 - a^2)}$.
Point charges $q_1 = 2 c and q_2 = –1 c$ are kept at points x = 0 and x = 6 respectively. Electrical potential will be zero at points .....
Potential will be zero at two points $ At internal point (M ) = {1 \over 4 \pi \varepsilon} \left[ { 2 \times 10^{-6} \over (6 - l ) }+ { ( -1 \times 10^{-6} )\over l } \right] = 0 \Rightarrow l = 2 $ So distance of M from origin; x = 6 -2 = 4 At exterior point (N ) $ {1 \over 4 \pi \varepsilon} \left[ { 2 \times 10^{-6} \over (6 - l ) }+ { ( -1 \times 10^{-6} \over l } \right] = 0 \Rightarrow l' = 6 $
Two thin wire rings each having a radius R are placed at a distance d apart with their axes coinciding. The charges on the two rings are +q and –q. The potential difference between the centres of the two rings is ....
N identical drops of mercury are charged simultaneously to 10 volt. when combined to form one large drop, the potential is found to be 40 volt, the value of N is ......
To find the number of identical mercury drops that combine to form one large drop, we use the relationship between the potential and the radius. When N drops combine, the total volume remains the same, so if each drop has radius r, the large drop will have radius $R = rN^{1/3}$. Potential V is proportional to the charge and inversely proportional to the radius. Given initial potential V = 10V and final potential 40V, we set up the equation $10N^{1/3} = 40$, which simplifies to $N^{1/3} = 4$, giving N = 64. Therefore, the closest option is 8, making option o3 the correct answer.
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.