Question

One of the hazards facing humans in space is space radiation: high-energy charged particles emitted by the sun. During a solar flare, the intensity of this radiation can reach lethal levels. One proposed method of protection for astronauts on the surface of the moon or Mars is an aligned of large, electrically charged spheres placed high above areas where people live and work. The spheres would produce a strong electric field E S to deflect the charged particles that make up space radiation. The spheres would be similar in construction to a Mylar balloon, with a thin, electrically conducting layer on the outside surface on which a net positive or negative charge would be placed. A typical sphere might be 5 m in diameter. Suppose that to repel electrons in the radiation from a solar flare, each sphere must produce an electric field E S of magnitude 1\( \times \) 10⁶ N/C at 25 m from the center of the sphere. What is the magnitude of E S just outside the surface of such a sphere? (a) 0; (b) 10⁶ N/C; (c) 10⁷ N/C; (d) 10⁸ N/C.

Short answer

The electric field outside the surface of each sphere is \({10^8}\;{\rm{N}}/{\rm{C}}\) and the option (d) is correct.

Step-by-step solution

Identification of given data

The magnitude of electric field outside the surface of each sphere is\(E = 1 \times {10^6}\;{\rm{N}}/{\rm{C}}\)

The diameter of each sphere is\(d = 5\;{\rm{m}}\)

The distance for electric field from the centre of each sphere is \(D = 25\;{\rm{m}}\)

Conceptual Explanation

The number of excess electron is found by calculating the charge due to electric field on the outside of sphere.

Determination of electric field outside the surface of sphere

The electric field just outside the surface of each sphere is given below:

\(E = \frac{{kQ}}{{{D^2}}}\)

Here,\(k\)is Coulomb’s constant and its value is\(9 \times {10^9}\;{\rm{N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}\).

Substitute all the values in the above equation.

\(\begin{aligned}1 \times {10^6}\;{\rm{N}}/{\rm{C}} = \frac{{\left( {9 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right)Q}}{{{{\left( {25\;{\rm{m}}} \right)}^2}}}\\Q = 0.0694\;{\rm{C}}\end{aligned}\)

The electric field outside the surface of each sphere is given as:

\(\vec E = \frac{{kQ}}{{{{\left( {\frac{d}{2}} \right)}^2}}}\)

Substitute all the values in the above equation.

\(\begin{aligned}\vec E = \frac{{\left( {9 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right)\left( {0.0694\;{\rm{C}}} \right)}}{{{{\left( {\frac{{5\;{\rm{m}}}}{2}} \right)}^2}}}\\\vec E = 0.0999 \times {10^9}\;{\rm{N}}/{\rm{C}}\\\vec E \approx {10^8}\;{\rm{N}}/{\rm{C}}\end{aligned}\)

Therefore, the electric field outside the surface of each sphere is \({10^8}\;{\rm{N}}/{\rm{C}}\) and the option (d) is correct.