If radius of gaussian surface enclosing a charge is halved, why is electric flux constant since flux = edscos(theta) and - Physics - Electric Charges And Fields

Question

If radius of gaussian surface enclosing a charge is halved, why is
electric flux constant since flux = edscos(theta) and flux is
directly proportional to surface area?

Ashish Marandi · Accepted Answer

Dear student,
<img alt="" src=
"https://s3mn%20mnimgs%20com/img/shared/content_ck_images/ck_5ef5cfc24dfc0%20jpg"
style="width: 1023px; height: 534px;">Regards

If radius of gaussian surface enclosing a charge is halved, why is electric flux constant since flux = edscos(theta) and flux is directly proportional to surface area?