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55 th question in the attachment....................................... .

55 th question in the attachment.......................................
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Grade:12

3 Answers

Arun
25750 Points
4 years ago
Let the charges are q1,q2
so q1=q2=2*10^(-16)C
distance,d=20cm so radius,r=d/2=5cm=5*10^(-2)m
force F=(10^(-9)*2*10^(-16)*2*10^(-16))/(9*25*10^(-4))
            =1.77*10^(-39)N
Therefore electric field intensity, E=F/q
                                                         =8.88*10^(-24)volts
 
Vikas TU
14149 Points
4 years ago
Dear student 
Let A s the midpoint 
When charge is placed at A 
E1 = kq1q2/r + kq3q4/r 
E1 = 72*10^-4 J 
When charge is placed at B , 20 cm from both charge,
E2 = Kq1q2/r + kq3q4/r 
E2 = 36*10^-4 J 
Work done = E1 - E2 
= 3.6*  10^-3 J
 
aswanth nayak
100 Points
4 years ago
 
Dear student ,
 

Let A is the midpoint 
When charge is placed at A 
E1 = kq1q2/r + kq3q4/r 
E1 = 72*10^-4 J 
When charge is placed at B , 20 cm from both charge,
E2 = Kq1q2/r + kq3q4/r 
E2 = 36*10^-4 J 
Work done = E1 - E2 
= 3.6*  10^-3 J
 
Regards
 

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