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

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11 months ago

```							Let the charges are q1,q2so q1=q2=2*10^(-16)Cdistance,d=20cm so radius,r=d/2=5cm=5*10^(-2)mforce F=(10^(-9)*2*10^(-16)*2*10^(-16))/(9*25*10^(-4))            =1.77*10^(-39)NTherefore electric field intensity, E=F/q                                                         =8.88*10^(-24)volts
```
11 months 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
```
11 months 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
```
8 months ago
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