Please calculate the net electric flux thru the cylinder. Answer is (1)
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Electric flux = E x A .... where E = Electric field and A = area ...
area = 2πRh ( R = 0.07 m and , h =1m )
=> area = 2 x (22/7) x 0.07 x 1 = 0.44 m^2
and E = 250N/C (given)
Now , Electric flux = 250 x 0.44 = 110
=> Electric flux = 1.1 x 10^2 Nm^2/C ....
area = 2πRh ( R = 0.07 m and , h =1m )
=> area = 2 x (22/7) x 0.07 x 1 = 0.44 m^2
and E = 250N/C (given)
Now , Electric flux = 250 x 0.44 = 110
=> Electric flux = 1.1 x 10^2 Nm^2/C ....
ishika1427:
Hey there,
I hv asked a few more questions which are unanswered. Could you please solve them too
hi... i am really sorry.... i am actually in car rn going somewhere... but.. I'll surely try to do them tonight.... sorry
Oh, no problem. Thanks a lot :)
okay... and.. you're welcome! ^_^
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Electric flux = E x A .... where E = Electric field and A = area ...
area = 2πRh ( R = 0.07 m and , h =1m )
=> area = 2 x (22/7) x 0.07 x 1 = 0.44 m^2
and E = 250N/C (given)
Now , Electric flux = 250 x 0.44 = 110
=> Electric flux = 1.1 x 10^2 Nm^2/C ....
area = 2πRh ( R = 0.07 m and , h =1m )
=> area = 2 x (22/7) x 0.07 x 1 = 0.44 m^2
and E = 250N/C (given)
Now , Electric flux = 250 x 0.44 = 110
=> Electric flux = 1.1 x 10^2 Nm^2/C ....
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