Electrostatics

Get insights from 79 questions on Electrostatics, answered by students, alumni, and experts. You may also ask and answer any question you like about Electrostatics

Follow Ask Question
79

Questions

0

Discussions

4

Active Users

0

Followers

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

3 months ago

0 Follower 3 Views

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

3 months ago

0 Follower 3 Views

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

3 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

E n e t = 4 k q d 2 * 2 c o s ? 30 ? = q 3 π ε 0 d 2

New answer posted

3 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

6

Sol. Common potential after connection.

V common   = C 1 V 1 + C 2 V 2 C 1 + C 2 = 60 * 20 + 0 120 = 10 V o l t

 

Loss of energy = 1 2 C V 2 - 1 2 ( 2 C ) * V Common   2

= 1 2 * 60 * 10 - 12 * ( 20 ) 2 - 60 * 10 - 12 * ( 10 ) 2 = 60 * 10 - 12 ( 200 - 100 ) = 6000 * 10 - 12 = 6 n J

 

New answer posted

3 months ago

0 Follower 9 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution
 

 

New answer posted

3 months ago

0 Follower 11 Views

A
alok kumar singh

Contributor-Level 10

Enet = Eo/k
Enet = E_free - E_bound = qf/Aε? - qb/Aε?
Eo = qf/Aε?
So, (qf-qb)/Aε? = qf/ (kAε? )
qf - qb = qf/k
qb = qf (1 - 1/k)

New answer posted

3 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

If charge (-Q) at origin is replaced by (+Q), then electric field at the centre of the cube is zero. Thus, electric field at the centre of the cube is as if only (-2Q) charge is present at the origin.

E = 1 4 π ε 0 ( 2 Q ) ( 3 2 a ) 2 . 1 3 ( x ^ + y ^ + z ^ ) = | 2 Q 3 3 π ε 0 a 2 ( x ^ + y ^ + z ^ )

New answer posted

3 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

When connected in series, equivalent capacitance,

C 1 = C * C C + C = C 2            

When connected in parallel, equivalent capacitance

C2 = C + C = 2C

C 1 C 2 = C / 2 2 C = 1 4            

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 679k Reviews
  • 1800k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.