Due to a charge inside a cube the electric field is Ex = 600 x1/2, Ey = 0 Ez - Physics - Electric Charges And Fields

Question

Due to a charge inside a cube the electric field is E
x = 600 x 1 /2,
E y = 0
E z = 0 The charge inside the cube is
(approximately) :

(A) 600 µC (B) 60 µC (C) 7 µµC (D) 6
µµC

Pintu B. · Accepted Answer

Dear Student ,
Here in this case ,applying Gauss's theorem we can write that ,
ϕ=qε0 where q is the total charge enclosed and ε0 is the free space permittivity 
Now the electric flux along the XY and XZ plane is zero as electric field is along X axis only 
So , ϕinside=∫E·ds=E∫ds=600x·0·12=6xSimilarly ,ϕoutside=∫E·ds=E∫ds=2×600x·0·12=12xSo total electric flux is , ϕ=ϕoutside-ϕinside=6xNow , ϕ=qε0⇒q=6ε0x
Regards

Due to a charge inside a cube the electric field is E x = 600 x 1 /2, E y = 0. E z = 0. The charge inside the cube is (approximately) : (A) 600 µC (B) 60 µC (C) 7 µµC (D) 6 µµC

Due to a charge inside a cube the electric field is E _x = 600 x ¹ ^/2, E _y = 0.
E _z = 0. The charge inside the cube is (approximately) :

(A) 600 µC (B) 60 µC (C) 7 µµC (D) 6 µµC