1
JEE Main 2021 (Online) 25th July Evening Shift
+4
-1
If qf is the free charge on the capacitor plates and qb is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge qb an be expressed as :
A
$${q_b} = {q_f}\left( {1 - {1 \over {\sqrt k }}} \right)$$
B
$${q_b} = {q_f}\left( {1 - {1 \over k}} \right)$$
C
$${q_b} = {q_f}\left( {1 + {1 \over {\sqrt k }}} \right)$$
D
$${q_b} = {q_f}\left( {1 + {1 \over k}} \right)$$
2
JEE Main 2021 (Online) 25th July Morning Shift
+4
-1
A parallel plate capacitor with plate area 'A' and distance of separation 'd' is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as :

$$\varepsilon (x) = {\varepsilon _0} + kx$$, for $$\left( {0 < x \le {d \over 2}} \right)$$

$$\varepsilon (x) = {\varepsilon _0} + k(d - x)$$, for $$\left( {{d \over 2} \le x \le d} \right)$$
A
$${\left( {{\varepsilon _0} + {{kd} \over 2}} \right)^{2/kA}}$$
B
$${{kA} \over {2\ln \left( {{{2{\varepsilon _0} + kd} \over {2{\varepsilon _0}}}} \right)}}$$
C
0
D
$${{kA} \over 2}\ln \left( {{{2{\varepsilon _0}} \over {2{\varepsilon _0} - kd}}} \right)$$
3
JEE Main 2021 (Online) 16th March Morning Shift
+4
-1
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 $${3 \over 4}$$d, where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance (C0) is given by the following relation :
A
$$C' = {{3 + K} \over {4K}}{C_0}$$
B
$$C' = {{4 + K} \over {3}}{C_0}$$
C
$$C' = {{4K} \over {K + 3}}{C_0}$$
D
$$C' = {{4} \over {3 + K}}{C_0}$$
4
JEE Main 2021 (Online) 26th February Morning Shift
+4
-1
Consider the combination of 2 capacitors C1 and C2 with C2 > C1, when connected in parallel, the equivalent capacitance is $${{15} \over 4}$$ times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, $${{{C_2}} \over {{C_1}}}$$.
A
$${{15} \over {11}}$$
B
No Solutions
C
$${{29} \over {15}}$$
D
$${{15} \over {4}}$$
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