Answer :
Use the conservation of energy:
mg = Eq
q = mg / E
= ( 0.200* 10 ^ -3 kg ) (9.8 m/s^2 ) / 3000 N /C )
= 0.653 micro C
b)
mg = T + Eq
mg - T = E q
q = T-mg / E
= ( 4 * 10 ^ -3 N ) -( 0.2*10^ -3 kg ) ( 9.8 m/s^2 ) / 3000 N /C
= 1.26 micro C
Tension in the thread is 0, when
Mg = force due to electric field
.200* 10^-3*9.8 =q*3*10^3
Therefore, q= .200*10^-3*9.8/3*10^3
Net tension =4*10^-3N
i.e. mg-qE = 4*10^-3
.200*9.8*10^-3
The ball is in equillibrium if it is under the effect of three forces that is unde the tension of the thread , the pull of gravity on the ball= w=mg= 0.2*10^-3*9.8=1.96*10^-3 N and the coulumb force having charge x which is downward if the ball is positve and upward if the ball is negative. Fc=qE =x*3000N/C.
first case
T-mg-Fc=0 => T=0
mg=Fc
1.96*10^-3=x*3000
x=6.5*10^-7 C charge
second case
T= 4.00mN=4*10^-3 N
4*10^-3 -1.96*10^-3- x*3000=0
x=6.08*10^-7C
mg = Eq
q = mg / E
= ( 0.200* 10 ^ -3 kg ) (9.8 m/s^2 ) / 3000 N /C )
= 0.653 micro C
b)
mg = T + Eq
mg - T = E q
q = T-mg / E
= ( 4 * 10 ^ -3 N ) -( 0.2*10^ -3 kg ) ( 9.8 m/s^2 ) / 3000 N /C
= 1.26 micro C
Tension in the thread is 0, when
Mg = force due to electric field
.200* 10^-3*9.8 =q*3*10^3
Therefore, q= .200*10^-3*9.8/3*10^3
Net tension =4*10^-3N
i.e. mg-qE = 4*10^-3
.200*9.8*10^-3
The ball is in equillibrium if it is under the effect of three forces that is unde the tension of the thread , the pull of gravity on the ball= w=mg= 0.2*10^-3*9.8=1.96*10^-3 N and the coulumb force having charge x which is downward if the ball is positve and upward if the ball is negative. Fc=qE =x*3000N/C.
first case
T-mg-Fc=0 => T=0
mg=Fc
1.96*10^-3=x*3000
x=6.5*10^-7 C charge
second case
T= 4.00mN=4*10^-3 N
4*10^-3 -1.96*10^-3- x*3000=0
x=6.08*10^-7C