You visited us 1 times! Enjoying our articles? Unlock Full Access!

Capacitors

A large condu...

Question

A large conducting plate X having charge + Q distributed uniformly over area A is placed parallel near a large conducting uncharged plate Y of same area. Find the electric field at a point situated : (a) left to both the plates (b) between the plates (c) right to both the plates

Open in App

Solution

Dear Student,

$(a) t h e c h a r g e w i l l d i s t r i b u t e o n t h e c h a r g e d p l a t e o n b o t h t h e s i d e e q u a l l y Q / 2 . c h a r g e d e n s i t y σ = \frac{Q}{2 A} f i e l d o n l e f t E = \frac{σ}{ε_{0}} = \frac{Q}{2 A ε_{0}} (b) b e t w e e n t h e p l a t e s t h e p l a t e y h a s a i n d u c e c h a r g e o g - Q / 2 o n t h e s i d e o f x a n d + Q / 2 o n t h e s i d e o p p o s i t e t o X E = \frac{Q}{2 A ε_{0}} + \frac{Q}{2 A ε_{0}} = \frac{Q}{A ε_{0}} (c) o n t h e r i g h t E = \frac{Q}{2 A ε_{0}} R e g a r d s$

Suggest Corrections

Similar questions

X

Y

A

X

Q

X

P, Q

R

σ, σ

- σ

S

Q

R

E_{1}

E_{2}

F_{1}

F_{2}

I

I I

	List $I$	List $I I$
p.	Now, the middle plate shifts towards the third plate very slowly keeping $1^{s t}$ and $3^{r d}$ plates fixed. Initial and final situations shown in figure. Then	1. $\frac{E_{1}}{E_{2}} > 1$
q.	Three large conducting parallel plates each having same area are connected to the battery and the space between the $2^{n d}$ and $3^{r d}$ plates is filled with dielectric. Now, the dielectric is removed very slowly. Then	2. $\frac{E_{1}}{E_{2}} < 1$
r.	Three large conducting parallel plates each having same area and having charges $- Q$ , $2 Q$ and $- Q$ respectively initially. Now first plate shifts towards the left and $3^{r d}$ towards right keeping $2^{n d}$ fixed. Then	3. $\frac{F_{1}}{F_{2}} < 1$
s.	Three large conducting parallel plates each having same area, charge on $1^{s t}$ plate is $-$ $Q$ , on $2^{n d}$ plate is $Q$ and the third is uncharged. Now, the switch is closed. If $F_{1}$ is the force between plate 1 and 3 initially and $F_{2}$ is the force between them finally, then	4. $\frac{F_{1}}{F_{2}} = 1$

σ

Join BYJU'S Learning Program