Two identical spheres having charges of opposite sign attract each other with a force of 0.108n when separated by 0.5 m . The spheres are connected by a conducting wire, which when removed and thereafter they repel each other with a force of 0.036 N. What were the initial charges on the spheres?

 

27 drops of mercury are charged simultaneously to the same potential of 10 V . What will be the potential if all the charged drops are made to combine to form one large drop? Assume the dorps to be spherical. 

 

A cube of side b has a charge q at each of its vertices. Determine the electric field due to theis charge array at the centre of the cube

 

2 particles have equal masses of 5.0 g each ans opposite charges of 4 * 10-5 & -4 * 10-5 . they are released from rest witha a separation of 1m between them . Find the speeds of the paricles when the separation is reduced to 50cm.

 

ABC is a rt. angled triangle, where AB & BC are 25 cm and 60 cm respectively a metal sphere of 2 cm radius charged to a potential of 9 * 10 ^5 V is placed at B Find the amt. of work done in Carrying a +ve charge of 1C from C toA

 

3 concentric metallic shells A B C of radii a< b<  c  have surface charge densities +  - & +  respectively

1. Find the potentialof 3 shells A B C

2. If shells A & C are at the same potential , obtain the relation between radii a b c

 

 

Hi nasa_hs
2) since volume of mercury remains const radius of new drop

R^3=27*r^3

-->R=3r

charge on new drop=27q where q=charge on individual drops(27)

V=1/4pie e*q/r

v1=10=k(q/r)

v2=k(27q/R)=k(27q/3r)=9(kq/r)=9v1=90V ans

 

3)Electric field at centre of cube=0

Whenever charge distribution is symmetric about a pt the E at that pt due to charge distribution becomes 0

 

1)let the charges be q1 and q2

F=kq1q2/r^2

9*10^9 q1q2/25*10^-2 =.108

--> q1q2=300*10^-14=3*10^-12  ---1)

SINCE THE METTALIC SPHERES R IDENTICAL THE CHARGES GET DISTRIBUTED SUCH THAT BOTH 'VE =CHARGES AFTER REDISTRIBUTION

q1'=q2'=q

kq^2/r^2=.036

=> q^2=100*10^-14

q=10^-6=q1-q2/2

-->q1-q2=2*10^-6---2)

solving 1)and 2) q1 q2 (initial charges can be obtained)

 

5) (though u avent numbered the q)

CHARGE ON THE METAL SPHERE

1/4 *q/r=9*10^5

-->q=2*10^-6=2C

POTENTIAL AT C(Vc)

Vc=1/4 *q/r=9*10^3*2/60*10^-2

Vc=3*10^4V

SIMILARLY

Va=1/4 *q/r=9*10^3*2/25*10^-2

Va=7.2*10^4V

Work done= q(V)= 1*4.2+10^4

W=4.2*10^4J ANS

Equal force acts on both particles and as they have the same mass acclns r also same.

velocities will also b same

BY ENERGY CONSERVATION

-kq^2/r = -kq^2/r/2+2*1/2mv^2   where k=1/4=9*10^9

--->v=(kq^2/rm)^.5

Substituting the values

v comes to be 53.66m/s

Pls tell me if my solns r correct

PLS RATE ME!

Last one

1)The potential inside a charged metallic sphere is constant and equal to the potenial on its surface ie

V=1/4 q/r

q=charge=(4r^2) where  is surface charge density

Va=1/4 q1/a +1/4 q2/b+1/4 q3/c

q1=(4a^2)*  q2=(4b^2)* -  q3==(4c^2)* 

similarly Vb=1/4 q1/b +1/4 q2/b+1/4 q3/c

and Vc=1/4 q1/c+1/4 q2/c+1/4 q3/c

-----------------------------------------------

2)Va=Vc

/(a-b+c)=/(a^2/c-b^2/c+c)

ac-bc+c^2=a^2-b^2+c^2

c(a-b)=(a+b)(a-b)

------> c=a+b

Pls tell me if iam correct