Chapter 21: Q15P (page 898)

The center of a thin paper cube in outer space is located at the origin. Each edge is 10 cm long. The only other objects in the neighborhood are some small charged particles whose charges and position at this instant are following: +9 nC at (4,2,-3) cm, -6 nC at (1,-3,2) cm, +7 nC at (15,0,4) cm, +4 nC at (-1,3,2) cm and -3 nC at (-2,12,-3) cm. The electric flux on the five of the six faces of the cube totals 564 V m. What is the flux on the other face?

Short Answer

Expert verified

The electric flux through other face of cube is $227 V . m$ .

Step by step solution

Identification of given data

The electric flux on the five faces of cube is $ϕ_{5} = 564 V \cdot m$

The first charge inside the cube is $Q_{1} = 9 nC$

The second charge inside the cube is $Q_{2} = - 6 nC$

The third charge inside the cube is $Q_{3} = 4 nC$

Conceptual Explanation

The charges that are lying inside the cube can only contribute in the flux of the cube and those lying outside the cube do not contribute in the total flux.

Determination of electric flux on the sixth face of cube

Apply the Gauss’s law to find the total amount of electric flux of cube:

$ϕ = \frac{Q_{1} + Q_{2} + Q_{3}}{ε_{0}}$

Here, $ε_{0}$ is the permittivity of box and its value is $8.854 \times 10^{- 12} C^{2} / N \cdot m^{2}$ .

Substitute all the values in the above equation.

$\begin{array}{rcl} ϕ & = & \frac{9 nC + (- 6 nC) + 4 nC}{8.854 \times 10^{- 12} C^{2} / N \cdot m^{2}} \\ ϕ & = & \frac{(7 nC) (\frac{10^{- 9} C}{1 nC})}{8.854 \times 10^{- 12} C^{2} / N \cdot m^{2}} \\ ϕ & = & 791 V \cdot m \end{array}$