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Two spheres with uniform surface charge density, one with a radius of 7.4 cm and the
other with a radius of 5.0 cm, are separated by a center-to-center distance of 38 cm. The
spheres have a combined charge of +55μC and repel one another with a force of 0.62 N.
Assume that the charge of the first sphere is greater than the charge of the second sphere.

Question 1.
What is the surface charge density in the sphere of radius 7.4?

Question 2
What is the surface density on the 2nd sphere?

EXPRESS ANSWER USING TWO SIGNIFICANT FIGURES.

Answer :

onyebuchinnaji

(a) The surface charge density in the sphere of radius 7.4 is 0.0322 C/m³.

(b) The surface density on the 2nd sphere is  3.48  x 10⁻⁴ C/m³.

Total charge of the spheres

F = kq₁q₂/r²

Fr² = kq₁q₂

q₁q₂ = Fr²/k

where;

  • r is distance between the charges
  • k is Coulomb's constant

q₁q₂ = (0.62 x 0.38²) / (9 x 10⁹)

q₁q₂ = 9.95 x 10⁻¹² C

q₂ = 9.95 x 10⁻¹² C/q₁

From the question;

q₁ + q₂ = 55 x 10⁻⁶

q₁ + 9.95 x 10⁻¹² /q₁ = 55 x 10⁻⁶

q₁² + 9.95 x 10⁻¹²  =  55 x 10⁻⁶q₁

q₁² - 55 x 10⁻⁶q₁  + 9.95 x 10⁻¹²   = 0

solve the quadratic equation using formula method;

q₁ = 5.48 x 10⁻⁵ C

q₂ = 55 x 10⁻⁶ C - 5.48 x 10⁻⁵ =  1.82 x 10⁻⁷ C

Volume of the first sphere

V1 = ⁴/₃πr³

V1 = (⁴/₃ π)(0.074)³ =  1.7 x 10⁻³ m³

Surface charge density = (5.48 x 10⁻⁵ C) / (1.7 x 10⁻³ m³) = 0.0322 C/m³

Volume of the second sphere

V2 = (⁴/₃ π)(0.05)³ = 5.236 x 10⁻⁴ m³

Surface charge density = ( 1.82 x 10⁻⁷ C) / (5.236 x 10⁻⁴ m³) = 3.48  x 10⁻⁴ C/m³

Thus, the surface charge density in the sphere of radius 7.4 is 0.0322 C/m³.

The surface density on the 2nd sphere is  3.48  x 10⁻⁴ C/m³.

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