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## Science Sample Papers - 2022 ### UNIT-1 **A.** We know the Dielectric constant ($K$) is: $K = \frac{E_1}{E_2} = \frac{E_1}{E_1 - E_2}$ **6** The electric field intensity between the plates of a parallel plate capacitor decreases by $20\%$ when a dielectric slab is inserted between the plates....

## Science Sample Papers - 2022 ### UNIT-1 **A.** We know the Dielectric constant ($K$) is: $K = \frac{E_1}{E_2} = \frac{E_1}{E_1 - E_2}$ **6** The electric field intensity between the plates of a parallel plate capacitor decreases by $20\%$ when a dielectric slab is inserted between the plates. Find the dielectric constant of the dielectric. **A.** As per question: * if $E_1 = 100$ * $E_2 = (100-20) = 80$ So $D.E.C = \frac{E_1}{E_1 - E_2} = \frac{100}{100 - 80} = 5$ **7.** Find the effective capacitance of the group of capacitors across AB; Fig. 3.14. **8.** Calculate the equivalent capacitance between A and B of the figure shown. [Fig 3.15]. **A.** A close study of the Fig 3.14 reveals that it is a parallel grouping of 4 capacitors. So, $C_p = C_1 + C_2 + C_3 + C_4 = 4C$ **A.** Capacitance 1.5µF and 1.5µF are in parallel. Their equivalent capacitance is 1.5 + 1.5 = 3µF Which is in series with two other capacitances 3µF each. Hence: