Two small spheres each carrying a charge $q$ are placed $r$ metre apart. If one of the spheres is taken around the other one in a circular path of radius $r$, the work done will be equal to
AForce between them $ \times \,r$
BForce between them $ \times \,2\pi r$
CForce between them $/\,2\pi r$
D
Zero
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D
Zero
d (d)The force is perpendicular to the displacement.
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Two opposite and equal charges $4 \times {10^{ - 8}}\,coulomb$ when placed $2 \times {10^{ - 2}}\,cm$ away, form a dipole. If this dipole is placed in an external electric field $4 \times {10^8}\,newton/coulomb$, the value of maximum torque and the work done in rotating it through $180^\circ $ will be
The plates of a parallel plate capacitor of capacity $50\,\mu C$ are charged to a potential of $100\;volts$ and then separated from each other so that the distance between them is doubled. How much is the energy spent in doing so
Four identical charges $ + \,50\,\mu C$ each are placed, one at each corner of a square of side $2\,m$. How much external energy is required to bring another charge of $ + \,50\,\mu C$ from infinity to the centre of the square......$J$ $\left( {{\rm{Given}}\frac{{\rm{1}}}{{{\rm{4}}\pi {\varepsilon _{\rm{0}}}}} = 9 \times {{10}^9}\,\frac{{N{m^2}}}{{{C^2}}}} \right)$
A parallel plate capacitor has a dielectric slab of dielectric constant $K$ between its plates that covers $1 / 3$ of the area of its plates, as shown in the figure. The total capacitance of the capacitor is $C$ while that of the portion with dielectric in between is $C _1$. When the capacitor is charged, the plate area covered by the dielectric gets charge $Q_1$ and the rest of the area gets charge $Q_2$. Choose the correct option/options, igonoring edge effects.
$(A)$ $\frac{E_1}{E_2}=1$ $(B)$ $\frac{E_1}{E_2}=\frac{1}{K}$ $(C)$ $\frac{Q_1}{Q_2}=\frac{3}{K}$ $(D)$ $\frac{ C }{ C _1}=\frac{2+ K }{ K }$
A parallel plate capacitor is charged to a potential difference of $50\, V$. It is discharged through a resistance. After $1$ second, the potential difference between plates becomes $40 \,V$. Then