A dielectric slab of thickness $d$ is inserted in a parallel plate capacitor whose negative plate is at $x = 0$ and positive plate is at $x = 3d$. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from $0$ to $3d$
IIT 1998, Medium
Download our app for free and get started
(d) Even after introduction of dielectric slab, direction of electric field will be perpendicular to the plates and directed from positive plate to negative plate.
Further, magnitude of electric field in air $ = \frac{\sigma }{{{\varepsilon _0}}}$
Magnitude of electric field in dielectric $ = \frac{\sigma }{{K{\varepsilon _0}}}$
Similarly electric lines always flows from higher to lower potential, therefore, electric potential increases continuously as we move from $x = 0$ to $x = 3d$.
Download our app
and get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
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......$J$ $\left( {{\rm{Given}}\frac{{\rm{1}}}{{{\rm{4}}\pi {\varepsilon _{\rm{0}}}}} = 9 \times {{10}^9}\,\frac{{N{m^2}}}{{{C^2}}}} \right)$
Two large vertical and parallel metal plates having a separation of $1 \ cm$ are connected to a $DC$ voltage source of potential difference $X$. A proton is released at rest midway between the two plates. It is found to move at $45^{\circ}$ to the vertical $JUST$ after release. Then $X$ is nearly