How does the voltage across the 9 pF and 11 PF capacitors? The voltage across th

Question

How does the voltage across the 9 pF and 11 PF capacitors? The voltage across the 9 pF capacitor is largest. The voltage across the 11 pF capacitor is largest. How does the charge on the 9 pF and 11 Pf capacitors compare? The charge on the 9 pF capacitor is largest. The charge on the 11 pF capacitor Is largest The charge is the Mine on the 9 pF und 11 pF capacitors. Two large metal plates carry equal and opposite charges spread over their surfaces, ax shown. Which statements about these plates arc correct? Choose all that apply. The electric potential at point a is equal to the potential at point b. The electric field strength at point a is equal to the field strength at point h. If a proton is released al point a ill will move with constant velocity toward point b. Three equal point charges are held in place as shown to his right. What is his direction of the net force on Q1? Left Right Up Down A parallel-plate capacitor having circular plates of radius R separated by a distance d is held at a fixed potential difference. If the plates are moved closer together the capacitance increases the capacitance decreases the capacitance stays the same

Explanation / Answer

Q6.

electric field due to a charged surface=sigma/(2*epsilon)

where sigma is the charge density.

as potential=integration of E*dx

where x is the distance

potential at a distance of x=sigma*x/(2*epsilon)

so if charge density on left plate is sigma, charge density on right plate is -sigma.

as no distance values are given , for simplicity, we can assume distance between the plates to be 3*d, distance of a from left plate to be d, distance of right plate from b to be d and distance between a and b be d.

potential at a=(sigma*d/(2*epsilon))-(sigma*2*d/(2*epsilon))

=-sigma*d/(2*epsilon)

potential at b=(sigma*2*d/(2*epsilon))-(sigma**d/(2*epsilon))

=sigma*d/(2*epsilon)

so potential at b is greater than potential at a.

(even if distance values are not as assumed, being at a larger distance from positive plate will give higher potential to a point.

in the current case,field strength does not depend upon distance.

hence field strength is equal for both a and b.

as electric field exist, it will exert a force on the proton at a and the proton will have an acceleration.

so it cant have constant velocity.

hence option c is correct.

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