Your answer is partially correct. Try again. A slab of copper of thickness b 1.9

Question

Your answer is partially correct. Try again. A slab of copper of thickness b 1.90 mm is thrust into a parallel-plate capacitor of plate area A 1.60 cm2 and plate separation d-5.00 mm, as shown in the figure; the slab is exactly halfway etween the plates. (a) what is the capacitance after the slab is introduced? (b) If a potential difference A = 65.0 V is maintained between the plates, what is the ratio of the stored energy before to hat after the slab is inserted? (c) How much work is done on the slab as it is inserted? (d) Is the slab sucked in or must it be pushed in? Units F (a) Number T 4.567e-13 (b) NumberTo.62 (c) Number TT5.6411e-12 (d) pushed in Units

Explanation / Answer

According to the concept of the electric potential and capacitance

Work done W=1/2cv^2

Given that

Thickness b=1.9mm

Distance between the plates d=5mm

Potential difference between the plates after the slab placed V=65 v

Capacitance after slab placed C=4.567*10^-13F

Now we find the work done on the slab

Work done w=1/2*4.567*10^-13*65^2

=9.65*10^-10 J

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