Question : Electric field in the given figure is directed along +X direction and given by Ex=5Ax+2B, where E is in NC-1 and x is in metre, A and B are constants with dimensions. Taking A=10 NC-1m-1 and B=5 NC-1m-1. Calculate the electric flux through the cube.
Doubt by Aditya
Solution :
Ex=5Ax+2B
A = 10 NC-1m-1
B=5 NC-1m-1
B=5 NC-1m-1
Side of the cube (l) = 10 cm
l = 10×10-2 m
l = 10-1 m
l = 10×10-2 m
l = 10-1 m
Area of each face of the cube
= area of square
= (side)2
= l2
= (10×10-2)2
= 100×10-4 m2
= 10-2 m2
= area of square
= (side)2
= l2
= (10×10-2)2
= 100×10-4 m2
= 10-2 m2
Here the electric field is only along the x-axis, hence the electric flux will be only along the face M and N of the cube. The flux through the other faces of the cube will be zero..
Electric flux along the face M of the cube
ΦM = EScosθ
ΦM = [5A(0)+2B][l2]cos180°
ΦM = [2B][l2](-1)
ΦM = -2Bl2
ΦM = EScosθ
ΦM = [5A(0)+2B][l2]cos180°
ΦM = [2B][l2](-1)
ΦM = -2Bl2
Electric flux along the face N of the cube
ΦN = EScosθ
ΦN = EScosθ
ΦN = [5A(l)+2B][l2]cos0°
ΦN = [5Al+2B][l2](1)
ΦN = 5Al3+2Bl2
Total flux through the cube
= ΦM + ΦN
= -2Bl2 + 5Al3+2Bl2ΦN = [5Al+2B][l2](1)
ΦN = 5Al3+2Bl2
Total flux through the cube
= ΦM + ΦN
= 5Al3
= 5(10)(10-1)3
= 50×10-3 Nm2C-1
= 5×10-2 Nm2C-1
= 5×10-2 Nm2C-1
Hence the total electric flux through the cube will be 5×10-2 Nm2C-1
