a) Two positive charges of7.61C eachare 768 cm apart. Find the electric field mi

Question

a) Two positive charges of7.61C eachare

768 cm apart.

Find the electric field midwaybetween

them.

1. E = 4973.19 N/C

2. E = 0 N/C

3. E = 52649 N/C

4. E = 7562.78 N/C

5. E = 8947.77 N/C

6. E = 20383.9 N/C

7. E = 13296.6 N/C

8. E = 30719.4 N/C

9. E = 9276.7 N/C

10. E = 12691.3 N/C

b) What is the magnitude of theelectric field if

one charge is positive and theother negative,

both of magnitude7.61C?

1. E = 5550.56 N/C

2. E = 6158.9 N/C

3. E = 6625.05 N/C

4. E = 7839.69 N/C

5. E = 4137.94 N/C

6. E = 3915.6 N/C

7. E = 9276.7 N/C

8. E = 18293.3 N/C

9. E = 0 N/C

10. E= 52327.8 N/C

Explanation / Answer

a). as both the charge are +ve, electric field at the center due to each charge is in oppositedirection, net electric field is 0, option 2 is correct. b) E1=kq/r2=8.9*109*7.6*10-6/7.682=1146.7 so electric field is E=2E1=2315N/C.

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