Question

Suppose that earth has a surface charge density of 1 electron/metre? Calculate earth's
potential and electric field just outside earth's surface. Radius of earth 6400 km.
​

Answer 1

Answer

Consider earth as a big conducting sphere with σ=−1.6×10

−19

Cm

−2

potential of a conducting sphere=kσ4πR

=9⋅10

9

×−1.6⋅10

−19

×4π×6400⋅10

3

≈−0.12V

ㅗㅐㅔㄷ ㅑㅅ ㅗㄷ ㅣㅔ ㅛㅐㅕ

Answer 2

Answer:

Od Earth has surface charge density of 1 electron/m² , then Electric potential is 1808.64 x 10¹⁴ V and Electric Field is 28.26 x 10 ⁹ V/m

Explanation:

Given that :

• Surface charge density of Earth = 1 electron/ m²
• Radius of Earth= 6400 km = 64 x 10⁵ m.

Solution :

• Area of Earth = πr² = 3.14 x ( 64 x 10⁵ )² m²

= 12861.44 x 10¹⁰ m²

• Surface charge on Earth, q = 12861.44 x 10¹⁰ electrons
• We know that electric potential at surface of a sphere due to uniform charged distribution is equal to electric potential when charge is supposed to be at the centre of sphere.
• Electric potential at the surface of Earth is given by:

$V = k \frac{q}{r} \\ = 9 \times {10}^{9} \times \frac{12861.44 \times {10}^{10} }{64 \times {10}^{5} } \\ = 1808.64 \times {10}^{14} \: V$

• Electric field at the surface is given by :

$Electric field, E = \frac{V}{r} \\ = \frac{1808.64 \times {10}^{14} }{ {64 \times {10}^{5}} } \\ = 28.26 \times {10}^{9} \frac{V}{m}$

• Hence, Electric potential is 1808.64 x 10¹⁴ V and Electric Field is 28.26 x 10 ⁹ V/m .