A parallel plate capacitor with air between the plate has a capacitance of 15 pF. The separation between the plate becomes twice and the space between them is filled with a medium of dielectric constant 3.5. Then the capacitance becomes (x/4) pF. The value of x is

Question

A parallel plate capacitor with air between the plate has a capacitance of 15 pF. The separation between the plate becomes twice and the space between them is filled with a medium of dielectric constant 3.5. Then the capacitance becomes (x/4) pF. The value of x is  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

C 0=(∈0 A / d )=15 pF C =( K ∈0 A /2 d )=(3.5/2) × 15 pF =(105/4) pF

Q. A parallel plate capacitor with air between the plate has a capacitance of $15 pF$ . The separation between the plate becomes twice and the space between them is filled with a medium of dielectric constant $3.5$ . Then the capacitance becomes $\frac{x}{4} pF$ . The value of $x$ is _______

$C_{0} = \frac{\in _{0} A}{d} = 15 pF$
$C = \frac{K \in _{0} A}{2 d} = \frac{3.5}{2} \times 15 pF = \frac{105}{4} pF$

Q. A parallel plate capacitor with air between the plate has a capacitance of 15pF. The separation between the plate becomes twice and the space between them is filled with a medium of dielectric constant 3.5. Then the capacitance becomes 4x​pF. The value of x is _______

C0​=d∈0​A​=15pF C=2dK∈0​A​=23.5​×15pF=4105​pF

Q. A parallel plate capacitor with air between the plate has a capacitance of $15 pF$ . The separation between the plate becomes twice and the space between them is filled with a medium of dielectric constant $3.5$ . Then the capacitance becomes $\frac{x}{4} pF$ . The value of $x$ is _______

$C_{0} = \frac{\in _{0} A}{d} = 15 pF$
$C = \frac{K \in _{0} A}{2 d} = \frac{3.5}{2} \times 15 pF = \frac{105}{4} pF$