(a) Two plates (20 cm x 3.0 cm) are separated by a 1.0 mm air gap forming a capacitor. Assume dielectric constant () for air is 1.000. The capacitor is connected to a 12 V battery. i) Determine the capacitance of the parallel plate capacitor. ii) What is the charge on each plate? (b) Three particles of charge 50 C, 27 C and –70 C are placed in a straight line with the 27 C charge in the middle and the 50 C charge to its left. The charge in the middle is 0.30 m from each of the other two charges. Calculate the magnitude and direction of net force on the 27 C charge due to the other two charges.

Question

Arun · Answer

C= d ϵ o ​ A ​ According to the values given C=1.8×10 −11 F Now Q=CV Hence charge on the positive plate will be 1.8×10 −9 C (a) Dielectric constant, k = 6 C=1.8×10 −11 F new capacitance, C ′ =kC=108 pF V=100volts q ′ =V×C ′ =1.08×10 −8 C V remains the same. (b) C ′ =kC=6×1.8×10 −11 =108pF q remains the same. q=1.8×10 −9 C V ′ =q/C ′ =16.67 V

Vikas TU · Answer

Dear student The above explanation is not properly explained Please ignore this We are sorry for inconvenience. Good Luck Cheers

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