JEE MAIN - Physics (2019 - 8th April Morning Slot - No. 26)
Voltage rating of a parallel plate capacitor is
500V. Its dielectric can withstand a maximum
electric field of 106 V/m. The plate area is
10–4 m2. What is the dielectric constant is the
capacitance is 15 pF?
(given $$\varepsilon $$0 = 8.86 × 10–12 C2/Nm2)
8.5
4.5
3.8
6.2
Explanation
A = 10–4 m2
Emax = 106 V/m
C = 15 $$\mu $$F
$$C = {{k{\varepsilon _0}A} \over d};{{Cd} \over {{\varepsilon _0}A}} = k$$
$$k = {{15 \times {{10}^{ - 12}} \times 500 \times {{10}^{ - 6}}} \over {8.86 \times {{10}^{ - 12}} \times {{10}^4}}}$$
= $${{15 \times 5} \over {8.86}} = 8.465$$
k $$ \approx $$ 8.5
Emax = 106 V/m
C = 15 $$\mu $$F
$$C = {{k{\varepsilon _0}A} \over d};{{Cd} \over {{\varepsilon _0}A}} = k$$
$$k = {{15 \times {{10}^{ - 12}} \times 500 \times {{10}^{ - 6}}} \over {8.86 \times {{10}^{ - 12}} \times {{10}^4}}}$$
= $${{15 \times 5} \over {8.86}} = 8.465$$
k $$ \approx $$ 8.5
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