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OpenStudy (anonymous):

how can we get electric field from electric potential?

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OpenStudy (anonymous):

\[V=E.d\]

OpenStudy (anonymous):

too*there is one eq of maxwell too but i dont remember

OpenStudy (anonymous):

\[Vab=Va-Vb=\int\limits_{a}^{?b}Edl\]

OpenStudy (anonymous):

from that formula

OpenStudy (anonymous):

your E is a constant in thah line?

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OpenStudy (anonymous):

yep

OpenStudy (anonymous):

take it off \[E.\int\limits_{a}^{2.b}dl=E(2.b-a)\]

OpenStudy (anonymous):

where is electic potential?

OpenStudy (anonymous):

you said Vab=E(2.b-a)

OpenStudy (anonymous):

oh i get it thanks.

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OpenStudy (anonymous):

A point charge Q is located at the center of a cube of side L. In addition, six other identical point charges q inside the cube are positioned symmetrically around Q, as shown in Fig. 4. Determine the electric flux through one face of the cube.

OpenStudy (anonymous):

could you put the pic?

OpenStudy (anonymous):

OpenStudy (anonymous):

that is the picture

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