Physics Electrostatic Potential and Capacitance

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New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

In air F = 1 4 π 0 q 1 q 2 r 2  

In medium F ' = 1 4 π ( k 0 ) q 1 q 2 r 2  

F ' = F 0 K

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

C = 30/d

d

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a month ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

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A
alok kumar singh

Contributor-Level 10

Direction of E is perpendicular to equipotential surface.

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a month ago

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

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alok kumar singh

Contributor-Level 10

C 1 = 1 μ F V 1 = 100 V ? U 1 = 1 2 * 1 * 10 - 6 * 100 2 = 5 * 10 - 3 J  

C 2 = 1 μ F , common potential V = C 1 V 1 C 1 C 2 = 100 2 = 50 V

Final electrostatic energy stored in both the capacitors.

= 1 2 C 1 + C 2 V 2 = 1 2 * 2 * 50 2 * 10 - 6

= 2.5 * 10 - 3 J

E n e r g y l o s t = 2.5 * 10 - 3

% l o s s o f e n e r g y = 2.5 * 10 - 3 5 * 10 - 3 * 100

= 50 %

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a month ago

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alok kumar singh

Contributor-Level 10

From work energy theorem

W s p + W E = ? K

- 1 2 k x 0 2 + Q E . x 0 = 0

? x 0 = 2 Q E k

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a month ago

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A
alok kumar singh

Contributor-Level 10

(Explanation can be elaborate)

For a region of constant electric potential

(1) The electric field may be uniform

(2) The electric field may be zero

(3) There can be no charge inside the region

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a month ago

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R
Raj Pandey

Contributor-Level 9

Before closing the switch

After closing the switch

So charge on C1  remains same, on C2 increases and on C3 and C4  decreases.

New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

Please find the answer the below

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