This BITSAT paper consists of electrostatics questions which are part of the BITSAT Physics section. Get BITSAT 2011 study material – BITSAT sample papers, BITSAT syllabus and start practicing for BITSAT 2011.
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. In the given circuit with steady current, the potential drop across the capacitor must be
. A quantity X is given by , where is the permittivity of free space, L is a length, is a potential difference and is a time interval. The dimensional formular for X is the same as that of
voltage
resistance
current
charge
. Consider a parallel plate capacitor of capacity (micro-farad) with air filled in the gap between the plates. Now one half of the space between the plates is filled with a dielectric of dielectric constant 4, as shown in the figure. The capacity of the capacitor changes to
. A capacitor is charged by a potential difference as shown in the figure. This charging battery is then removed and the capacitor is connected as in to an uncharged capacitor. What is the final potential difference (V) across the combination ?
. Two capacitors of capacitances and are charged to potentials and respectively. When they are connected in parallel the ratio of their respective charges will be
. Minimum numbers of and 250 V capacitors used to make a combination of and l000Vare
3
4
8
32
. In a parallel plate capacitor, the distance between the plates is d and potential difference across the plates is V. Energy stored per unit volume between the plates of capacitor is
. The equivalent capacitance of the combination of three capacitors, each of capacitance C shown in the given figure, between A and B is
. Charge Q is divided into two parts which are then kept some distance apart. The force between them will be maximum if the two parts are
e and (Q – e), where e = electronic charge
. A half ring of radius R has a charge of per unit length .The potential at the centre of the half ring is (Assuming potential at infinity is zero and )
. A point charge Q is moved along a circular path around another fixed point charge. The work done is zero
in all cases
only if Q returns to its starting point
only if the two charges have the same magnitude
only if the two charges have the same magnitude and opposite signs
. Electric potential of earth is taken to be zero, because earth is a good
insulator
conductor
semi-conductor
dielectric
. The pressure inside and outside a soap bubble of radius r are same. If the surface tension of the soap solution is T then the charge on the bubble will be
Zero
. Equal charges q are situated at the three corners of an equilateral triangle. The value of V and E at the centroid O will be
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