The election field in a region is given by $\vec E = (Ax + B)\hat i$ where $E$ is in $N\,C^{-1}$ and $x$ in meters. The values of constants are $A = 20\, SI\, unit$ and $B = 10\, SI\, unit$. If the potential at $x =1$ is $V_1$ and that at $x = -5$ is $V_2$ then $V_1 -V_2$ is.....$V$
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A parallel plate capacitor after charging is kept connected to a battery and the plates are pulled apart with the help of insulating handles. Now which of the following quantities will decrease?
Which one statement is correct ? A parallel plate air condenser is connected with a battery. Its charge, potential, electric field and energy are ${Q_o},\;{V_o},\;{E_o}$ and ${U_o}$ respectively. In order to fill the complete space between the plates a dielectric slab is inserted, the battery is still connected. Now the corresponding values $Q,\;V,\;E$ and $U$ are in relation with the initially stated as
In a parallel plate condenser, the radius of each circular plate is $12\,cm$ and the distance between the plates is $5\,mm$. There is a glass slab of $3\,mm$ thick and of radius $12\,cm$ with dielectric constant $6$ between its plates. The capacity of the condenser will be
For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant $K$ is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4} d$, where $'d'$ is the separation between the plates of parallel plate capacitor. The new capacitance $(C')$ in terms of original capacitance $\left( C _{0}\right)$ is given by the following relation
Assertion : If the distance between parallel plates of a capacitor is halved and dielectric constant is three times, then the capacitance becomes $6\,times$.
Reason : Capacity of the capacitor does not depend upon the nature of the material.
Two insulating plates are both uniformly charged in such a way that the potential difference between them is $V_2 - V_1 = 20\ V$. (i.e., plate $2$ is at a higher potential). The plates are separated by $d = 0.1\ m$ and can be treated as infinitely large. An electron is released from rest on the inner surface of plate $1. $ What is its speed when it hits plate $2?$
$(e = 1.6 \times 10^{-19}\ C, m_e= 9.11 \times 10^{-31}\ kg)$
As shown in the figure, two parallel plate capacitors having equal plate area of $200\,cm ^2$ are joined in such a way that $a \neq b$. The equivalent capacitance of the combination is $x \varepsilon_0 F$. The value of $x$ is $..........$.