A charge q is enclosed by a spherical surface of radius r. If the radius is reduced to half, how would the
electric flux through the surface change?
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HEYA GM ❤
¥ ELECTRIC FLUX :- q / Ë0 ( epsilon knot)
★ WHEN CHARGE q ; is placed in sphere of Radius :- R....
∆ Electric Flux :- q / Ë0 ( due to Whole sphere )
★ When Radius Decreased to R / 2 .... The flux is :::
∆| Electric Flux :- q / E0
∆| Electric Flux :- q / E0 » "Same as Above" .....
→ REASON :- FLUX IS INDEPENDENT OF RADIUS OF MATERIAL .... IT ONLY DEPENDS UPON....THE CHARGE ENCLOSED.... BY THE OBJECT DIVIDED BY E0...
TQ ❤ ❤
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