Physics Electrostatic Potential and Capacitance

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New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

When the two conducting spheres are connected by a wire, they will reach the same electric potential, V.
The total charge Q_total = 12µC + (-3µC) = 9µC. This total charge will redistribute.
Let the final charges be q? and q? + q? = 9µC.
The potential of a sphere is V = kq/r.
V? = V?
k q? /R? = k q? /R? ⇒ q? /R? = q? /R?
q? / (2R/3) = q? / (R/3) ⇒ q? /2 = q? ⇒ q? = 2q?
Substitute this into the charge conservation equation:
2q? + q? = 9µC ⇒ 3q? = 9µC ⇒ q? = 3µC.
Then, q? = 2 * 3µC = 6µC.
The final charges are 6µC and 3µC.

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

2 months ago

0 Follower 9 Views

V
Vishal Baghel

Contributor-Level 10

Heat loss; ΔH = U? - U? = 1/2 (C? / (C? +C? ) (V? -V? )²
= 1/2 * (5*2.5)/ (5+2.5) (220-0)²µJ
= 5/ (2*3) * 22*22*100*10? J
= 5*11*22/3 * 10? J = 1210/3 * 10? J = 1210/3 * 10? ³ J * 4 * 10? ²
According to questions
x/100 = 4*10? ²
So, x=4
Note: But given answer by JEE Main x=36

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

Before inserting slab
C_i = ε? A/d
After inserting dielectric slab
C_i = ε? lw/d
C_f = C? + C?
C_f = (Kε? A? /d) + (ε? A? /d)
C_f = 2C_i ⇒ (Kε? wx/d) + (ε? w (l-x)/d) = 2ε? lw/d
4x + l - x = 2l
x = l/3

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