IIT JEE , AIEEE Forum - Physics , Electricity - H.C VERMA- ELECTRIC FIELD AND POTENTIAL

sulekha_hi

Two identical pith balls, each having a charge 2 * 10^-7C and mass of 1000g are suspended from a common point by two insulating strings each 50 cm long. The balls are held at a separation 5 cm apart and then released. Find a) the electric force on one of the charged balls

b) the components of the resultant force on it along and perpendicular to the string

c) tension in the string

d) acceleration of one of the strings

I have solved all bits but i have some confusion in bit 2

plkease help me

raghu_21

its very easy...............

oac;is formed by strings;;so f/ac=mg/oc;;
it is balnced ..for more refeer..pradeep fundamental physicss..

sulekha_hi

Rahul, I know this is very easy i have also solved it but i have some concepts confused which is related to this question .So i want some more information with detailed concepts, if possible a diagram to explain components along and perpendicular to string. By the way thank u for your answer.

ppreyas

refer to modern's ABC of physics

phyana

hi sulekha
post ur doubt
let us analyze...............

imnotserious_2008

a) F=KQ^2/R^2 where R = 0.5 m

b) along the string it would b zero bcoz there is no acceleration in the string itself

perpendicular to string it wud be Fcos(theta) - mgsin(theta)

c) t= Fsin(theta) + mgcos(theta)

d) since F=ma

a= F/m

here F= resutant force perpendicular to the string.

find theta using hypotenuse law

cvramana

imnotserious_2008 your answer is partly correct. Acceleration of one of the balls will be in the resultant force direction, which is perpendicular to the string. So use the second answer to get the acceleration.

phyana

1 value of force is as explained by imnosterious2008
2 along the string tension acts.value of tention at equilibrium can be found using lames theorem(using the same find the equilibrium angle)
once u find the anggle, tention find the angle bet the force and line perp. to string. sum theses forces(cos componentof theforces )
3.find tetion as above
4the string dont have a const acc.......at any angle theta wit de vertical acc is cos of acc due to gravity

Vinodh.rsp

a) F=KQ^2/R^2 where R = 0.5 m

Can someone xplain this to me .plz

Vinodh

phyana

hey vinodh
f=(ele.feild)*charge
elcric feild due to one ball on other =k*q/(.5)^2
so the total force= k * q*q / (.5)^2
here .5 is the dstance of separation
force of one ball on the other is equl and opposite to the force on other on the one.
hope ugot it
post if doubts exist
rate my ans plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

Vinodh.rsp

Thanks for the reply

The balls are held at a separation 5 cm apart

The distance of separation is 5cm.How it'll change to 0.5 in the eqn?

The balls are held at a separation 5 cm apart and then released

If they are released they should be moving.so the force should change .How we can calculate the force just by using the initial distance of separation.

Thanks,

Vinodh

Arjun

the distance is .05 m.......not .5.........

did ya get it ?

since the force is changing ......we need to calculate the resultant of all forces.........inclding forces like mgsintheta............

so after calculting...the net resultant force......nd angle as said by

i'mnotsrious2008........we get da resultant force..........

hope u get it.....

phyana

hey vinodh
initially the system is not in equilibrium
then after some movement it attains equilibrium
when it attains equilibrium the resultant force acting on the particle wold be zero.using that u can find the ditance of separation using theta.all the above questions can be answered with the angle obstained
plz rate my answer

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