Electrostatic Potential and Capacitance

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New answer posted•a year ago
New answer posted•a year ago
New answer posted•a year ago
New answer posted•a year ago
S
a year ago

Battery is connected while dielectric is inserted so potential difference will be  remains same.

U i = 1 2 c V 2

U f = 1 2 K c V 2

⇒ Δ U = 1 2 ( K − 1 ) c V 2 = 1 2 × 1 × 2 0 0 × 1 0 − 9 × 2 0 0 2 = 4

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New answer posted•a year ago
S
a year ago

From observation, we can say that right of − 5 μ c charge, net electric force (or electric field) can be zero.

S o , k ( 5 μ c ) q x 2 = k ( 2 0 μ c ) q ( 5 + x ) 2

⇒ 1 x 2 = 4 ( 5 + x ) 2

⇒ 5 + x x = 2

x = 5 cm

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

C e q = C + C 1 . C 2 C 1 + C 2              

= ε 0 A 2 d + k 1 ε 0 A 2 d 2 . k 2 ε 0 A 2 d 2 k 1 ε 0 A 2 d 2 + k 2 ε 0 A 2 d 2              

= ε 0 A d ( 1 2 + k 1 k 2 k 1 + k 2 )  

 

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New answer posted•a year ago
S
a year ago

V A = 1 4 π ε 0 [ 1 a − 1 a 2 + s 2 ]

V B = − 1 4 π ε 0 [ 1 a − 1 a 2 + s 2 ]

V A − V B = 1 2 π ε 0 [ 1 a − 1 a 2 + s 2 ]

 

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New answer posted•a year ago
S
a year ago

for discharging RC – circuit

q = q 0 e − t τ         

l = d q d t = q 0 τ e − t / τ       

τ = C R = ( k ε 0 A d ) ( ρ d A ) [ ? l = d ]       

l m a x = q 0 ρ k ε 0 = 2 × 1 0 − 1 2 × 4 0 2 0 0 × 5 0 × 8 . 8 5 × 1 0 − 1 2

= 0.9 mA

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New answer posted•a year ago
S
a year ago

Assuming shell to be conducting

For  0 ≤ r < R q i n = 0 ,

E = 0

R ≤ r < a q i n = Q       

E = k Q r 2            

a ≤ r < b q i n = Q − Q = 0          

E = 0

r ≥ b q i n = Q

E = k Q r 2

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