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Physics 9 Online
OpenStudy (maheshmeghwal9):

Plz help: - If the case is |qA|>|qB|. Then why the electric field density is much more at (+)ve charge?

OpenStudy (maheshmeghwal9):

OpenStudy (maheshmeghwal9):

That is: -\[|^qA|>|^qB|.\]

OpenStudy (maheshmeghwal9):

Magnitude of charge of qA is more than that of qB.

OpenStudy (anonymous):

If the magnitude of charge is more then electric field density around it will be more

OpenStudy (maheshmeghwal9):

Thanx:)

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