Learn concept of electric flux definition with examples and equation. Practice electric flux formula and calculate electric flux through a cylinder.
Since,
Area c > Area b > Area a
Hence, flux through
c > b > a
Flux of Ring (c) is zero as its plane is oriented along the field so, no rain is passing through its area.
Electric flux is defined as the number of field lines passing through an area.
A c > b > a
B a = b = c
C a > b > c
D a > c > b
then total flux linked with the body will be zero.
then, Net flux= number of field lines leaving - number of field lines entering
Number of field lines passing through tilted surface (A) = Number of field lines passing through projection of area of surface.
\(\phi_{E_\perp'}=\phi_{E'}<\phi_E\)
\(A_{E_\perp'} < A_{E'} = A_E\)
A \(\phi_P>\phi_Q\)
B \(\phi_Q>\phi_P\)
C \(\phi_P=\phi_Q\)
D
By convention, area vector is always taken outwards / perpendicular to the plane as shown in figure.
Consider a ring placed in a uniform electric field, as shown in figure.
\(\phi_E=E(A\;cos\theta)\)
\(\phi_E=\vec E\cdot\ \vec A\)
\(\Delta\phi_E=E(\Delta A\;cos\theta)\)
or \(\Delta\phi_E=\vec E\cdot\Delta \vec A\)
where, \(\Delta\phi_E\) is the electric field through this element.
\(\phi_E=\sum\vec E\cdot\Delta\vec A\)
\(\phi_E=\int_s\vec E\cdot d\vec A\)
A 0 Nm2/C
B 36 Nm2/C
C 30 Nm2/C
D 400 Nm2/C
By convention, area vector is always taken outwards / perpendicular to the plane as shown in figure.
Consider a ring placed in a uniform electric field, as shown in figure.
\(\phi_E=E(A\;cos\theta)\)
\(\phi_E=\vec E\cdot\vec A\)
A 300 Nm2/C
B 3 Nm2/C
C 3000 Nm2/C
D 0
By convention area vector is always taken perpendicular and in outward direction to the surface.
Angle between \(\vec E\) and \(d\vec A_1\) is 180°.
\(\phi_1=\vec E\cdot d\vec A_1\)
\(\phi_1=E\;dA_1\;cos(180°)\)
\(\phi_1=-E\, dA_1\)
CONCLUSION: For any particular surface through which the electric field lines are entering, flux will be negative.
Angle between \(\vec E\) and \(d\vec A_2\) is 90°.
\(\phi_2=\vec E\cdot d\vec A_2\)
or, \(\phi_2=E\;dA_2\;cos(90°)\)
or, \(\phi_2=0\)
CONCLUSION: If electric field and area vector are perpendicular to each other, then flux through that surface is zero.
Angle between \(\vec E\) and \(d\vec A\) is 0°.
\(\phi_1=\vec E\cdot d\vec A\)
\(\phi_1=E\;dA\;cos0°\)
\(\phi_1=E\, dA\)
CONCLUSION: For any particular surface through which the electric fields are leaving, flux will be positive.
A –ve, –ve, –ve
B +ve, +ve, –ve
C +ve, +ve, +ve
D –ve, –ve, +ve
\(\phi_{E'_\perp}=\phi_{E'}<\phi_E\)
\(A_{E'_\perp} < A_{E'} = A_E\)
\(\phi=E(A\;cos\theta)\)
or, \(\phi=\) (Electric field) (projection of area perpendicular to electric field)
A \(E_0a^2\)
B \(E_0a^2\dfrac {\sqrt3}{2}\)
C \(E_0a\)
D \(E_0/a\)