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A particle A has charge +q and a particle B has charge +4 q with each of them having the same mass m. When allowed to fall from rest through the same electric potential difference, the ratio of their speed vAvB will become

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Important Questions on Electrostatics

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In an electron gun, electrons are accelerated through a potential difference of V volt. Taking electronic charge and mass to be respectively e and m, the maximum velocity attained by them is:
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In a cathode ray tube, if V is the potential difference between the cathode and anode, the speed of the electrons, when they reach the anode is proportional to: (Assume initial velocity =0)
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An electron of mass m and charge e is accelerated from rest through a potential difference V in a vacuum. The final speed of the electron will be
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Positive and negative point charges of equal magnitude are kept at 0,0,a2 and 0,0,-a2 , respectively. The work done by the electric field when another positive point charge is moved from -a, 0, 0 to 0, a, 0 is 
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If a positive charge is shifted from a low potential region to a high potential region, the electric potential energy
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An electron is accelerated by 1000 V, potential difference, its final velocity is:
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As per this diagram a point charge: +q is placed at the origin O. Work done in taking another point path charge -Q from the point A [co-ordinates o, a] another point B [co-ordinates a, o] along the straight path AB is:

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Two charges q1 q2 are placed 30 cm apart, as shown in the figure. A third charge q3 is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is q34πε0k, where k is:

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