Electric field in a region is given by vecE=-4 x hati+6 y hatj. The charge enclosed in the cube of side 1 m oriented as shown in the diagram is given by α ∈0. The value of α is

Question

Electric field in a region is given by vecE=-4 x hati+6 y hatj. The charge enclosed in the cube of side 1 m oriented as shown in the diagram is given by α ∈0. The value of α is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/f4e639824332cbde272c94c300140bc3-.png" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/ed458f657abac165102c1400adadeb8f-.png" /> φ=(6 y) Area -(4 x) Area =6 × 1 ×(1)2 -4 × 1 ×(1)2=2 therefore (q/ε0)=2 ⇒ q=2 ε0

Q. Electric field in a region is given by $E = - 4 x \hat{i} + 6 y \hat{j}$ . The charge enclosed in the cube of side $1 m$ oriented as shown in the diagram is given by $α \in_{0}$ . The value of $α$ is___

$ϕ = (6 y)$ Area $- (4 x)$ Area $= 6 \times 1 \times (1)^{2}$
$- 4 \times 1 \times (1)^{2} = 2$ therefore $\frac{q}{ε _{0}} = 2 \Rightarrow q = 2 ε_{0}$

Q. Electric field in a region is given by E=−4xi^+6yj^​. The charge enclosed in the cube of side 1m oriented as shown in the diagram is given by α∈0​. The value of α is___

ϕ=(6y) Area −(4x) Area =6×1×(1)2 −4×1×(1)2=2 therefore ε0​q​=2⇒q=2ε0​

Q. Electric field in a region is given by $E = - 4 x \hat{i} + 6 y \hat{j}$ . The charge enclosed in the cube of side $1 m$ oriented as shown in the diagram is given by $α \in_{0}$ . The value of $α$ is___

$ϕ = (6 y)$ Area $- (4 x)$ Area $= 6 \times 1 \times (1)^{2}$
$- 4 \times 1 \times (1)^{2} = 2$ therefore $\frac{q}{ε _{0}} = 2 \Rightarrow q = 2 ε_{0}$