Expression for an electric field is given by $\overrightarrow{ E }=4000 x^2 i \frac{ V }{ m }$ The electric flux through the cube of side $20 cm$ when placed in electric field (as shown in the figure) is ___$V cm$
The correct answer is 640
Flux \(=\vec{E}⋅\vec{A} \)
\(=4000(0⋅2)^2\frac{V}{m}⋅(0⋅2)^2m^2 \)
=4000×16×10−4Vm
=640 Vcm
\[ \Phi_E = \vec{E} \cdot \vec{A} \]
where \( \vec{E} \) is the electric field and \( \vec{A} \) is the area vector. The area of one face of the cube is:\[ A = (0.2 \, \text{m})^2 = 0.04 \, \text{m}^2 \]
Since the electric field is along the \( x \)-axis and the area vector is normal to the face of the cube, we have:\[ \Phi_E = E \times A = 4000 \times (0.2)^2 = 4000 \times 0.04 = 640 \, \text{Vcm} \]
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It is the property of subatomic particles that experiences a force when put in an electric and magnetic field.
It is a property associated with each point in space when charge is present in any form. The magnitude and direction of the electric field are expressed by E, called electric field strength or electric field intensity.
Electric charges are of two types: Positive and Negative. It is commonly carried by charge carriers protons and electrons.
Various properties of charge include the following :-
Two kinds of electric charges are there :-
When there is an identical number of positive and negative charges, the negative and positive charges would cancel out each other and the object would become neutral.