Physics Electrostatic Potential and Capacitance

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New answer posted

5 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

5 months ago

0 Follower 24 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

5 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

New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

Potential of centre, = V =
Vc = K (Σq)/R
Vc = K (0)/R = 0
Electric field at centre E_B = E_B = ΣE
Let E be electric field produced by each

charge at the centre, then resultant electric field will be
Ec = 0, since equal electric field vectors are acting at equal angle so their resultant is equal to zero.

New answer posted

5 months ago

0 Follower 10 Views

V
Vishal Baghel

Contributor-Level 10

Now, using junction analysis
We can say, q? + q? + q? = 0
2 (x - 6) + 4 (x - 6) + 5 (x) = 0
x = 36/11, q? = 36 (5)/11 = 180/11
q? = 16.36µC

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