IIT JEE , AIEEE Forum - Physics , Electricity

saanchi.arora

Three identical metal plates with large surface areas are kept parallel to each other. The leftmost plate is given a charge Q,the rightmost a charge -2Q and the middle one remains neutral. Find the charge appearing on the outer surface of each plate.

him26.89

it is -Q/2.

the charge on outer surface of both will b half of sum of both the charges.

umang

Here's the answer -

number all the 6 surfaces from first to last

charge on 1 = -q/2

2 = 3q/2

3 = -3q/2

4 = +3q/2

5 = -3q/2

6 = -q/2

just remember this = If there are n plates and charges given on each are q1,q2,q3,q4,... qn . Therefore , charge on outer surfaces of first and last plate is (q1+q2+q3+q4+q5+.......qn)/2

magiclko

Well these types of questions are done normally using gauss law, u can get that in H C Verma, in case if u dnt get even after seeing it, do knock back we are here for u

here's a shortcut to such questions,

find the total net charge q by summing all the given charges...(including signs)

and then distribute it equally to the outer most surfaces of the plates...then for the rest of the charge configuration, use charge conservation...

it will be like this in the given question

here total charge = -1Q

thrfore charge on each 1a and 3b = -Q/2 ........(1)

now using charge conservation for plates individually, 1b will have Q- (-Q/2) = 3Q/2 charge

as the plates are conducting an opposit charge will be induced on 2a, thrfore charge on 2a = -3Q/2

and since middle plate was initially uncharged, so charge on 2b= 3Q/2

similarly plate 3a will have charge -3Q/2

and 3b have charge -Q/2 (both by charge conservation and frm (1) )

ananth_patri

umang did it right....so did our form expert magiclko

umang

Thanks ananth !

It would be really nice if u could rate me !!

Thanks !!!

paridhi_aggarwal

yes !!
even i have the same answer as posted by manasi
the charge appearing on the outermost plates will be half the sum on the two plates

neeraj_agarwal_1990

And the facing sides will have charge same in magnitude but different in signs

magiclko

the answer is no doubt right, but still the proper method is the gauss law one

and also this method can be used only when none of th eplates are earthed, in that case u'll have to do it frm basics steps... so its necessary to knw that too

Ask & Discuss Questions with Community & Experts