Practice free Electrostatics (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
The electric potential V is given as a function of distance x (metre) by $V = (5x^2 + 10x – 9) $ volt. Value of electric field at x = 1 is .....
$ E = {-dv \over dx } = { -d \over dx } ( 5 x^2 + 10x -9 ) = -10x -10 $ $ \therefore E_(x-1) = - 10 \times 1 -10 = -20 { v \over m} $
A sphere of radius 1cm has potential of 8000 v, then energy density near its surface
will be ...
Energy density $ ue = { 1 / 2 } \varepsilon_o E^2 = {1/2 } 8.86 \times 10^{-12 } \times ( {v \over r} ) ^2 = 2.83 J/m^3 $
If a charged spherical conductor of radius 10cm has potential v at a point distant 5 cm from its centre, then the potential at a point distant 15cm from the centre will be .....
Potential inside the Sphere will be same as that on i ts Surface i .e. $v = V_{surface} ={q / 10} volt$ $ Vout = { q /15 } volt $ $ \therefore { vout / v } = { 2 /3 } \Rightarrow V_{out} = { 2 /3 } V $
The displacement of a charge Q in the electric field $ \bar E = e_1 \hat i + e_2 \hat j + e_3 \hat k $ is $ \bar r = a \hat i + b \hat j $ . The work done is
$ by using W = Q ( \bar E. |bar { \triangle r } ) $ $ W = Q [ ( e_1 \hat i + e_2 \hat j + e_3 \hat k ) . (a \hat i + b \hat j ) ] = Q (e_1 a + e_2 b ) $
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 }$
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