Two parallel plates, each having area A- 2100 cm are connected to the terminals

Question

Two parallel plates, each having area A- 2100 cm are connected to the terminals of a battery of voltage Vb 6 V as shown. The plates are separated by a distance d 0.41 cm. 1) What is Q, the charge on the top plate? 27.197 10n-10 C submit You currently have 2 submissions for this question. Only 10 submission are allowed. You can make 8 more submissions for this question. 2) What is U, the energy stored in the this cap J Submit 81.54 10A-10 You currently have 1 submissions for this question. Only 10 submission are allowed. You can make 9 more submissions for this question

Explanation / Answer

Given that

distance d=0.41cm

area A=2100 cm^2

vltage V=6v

now we find the capacitance of plates

capacitance C=eA/d=8.85*10^-12*2100*10^-4/41*10^-4

=453.3*10^-12 F

charge Q=VC=6*453.3*10^-12=2719.8*10^-12 C

now we find the energy sroed in the capacitor

the energy stored U=1/2cv^2=1/2*453.3*10^-12*6^2=8159.4*10^-12 J

now we find the energy stored in the new capacitor

capacitance C'=d1/d2*C

=41*10^-4/82*10^-4*453.3*10^-12=226.7 *10^-12

the energy stored in new capacitor U'=q^2/2C'=(2179.8*10^-12)^2/2*226.7*10^-12=10482.1*10^-12 J

now we find the electric field b/w the plates

the electric field E=V/d=6/41*10^-4=1463.4 N/c

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