Question

A uniform disc of charge has a total charge of 7.00x10-12 C and a radius of 10.0 cm.

A) Find the electric field at a point along its axis 20.0 cm from its center. (1.33 N/C away*)

B) Show that the electric field drops as 1/x2 as x approaches ?. Hint: remember that (1+x)n ~ 1+nx if x<<1)

Answer 1

A uniform disc of charge has a total charge of 7.00x10^-12 C and a radius of 10.0 cm.

Total charge, Q = 7 * 10^-12 C

Radius of the disc is, R = 10 cm = 0.1 m

k = 9 * 10⁹ N.m² / C² is the Coulomb's force constant

A) Electric field at a point along its axis 20.0 cm from its center.

x = 20 cm = 0.2 m

E = 2 k Q [1 - x /(x² + R²)^1/2] / R²

By plugging in the values, we get,

E = 2 * 9 * 10⁹ * 7 * 10^-12 * [1 - 0.2/(0.04+0.01)^1/2] / (0.1)²

E = 1.33 N/C

Direction:

Since the the disc has net positive charge, the electric field points away from the disc.

B) Show that the electric field drops as 1/x² as x approaches .

When x >> R, equation for the electric field becomes,

E = 2 k (Q/ R²) * [1 - x /(x² + R²)^1/2]

We use the approximation, (1+x)ⁿ ~ 1+nx if x<<1

E = 2 k (Q/ R²) * [1- x / (x² + 1/2R²)]

E = 2 k (Q/ R²) * [x² + 1/2R² - x / (x² + 1/2R²)]

Since x approaches , x² x . So,

E = 2 k (Q/ R²) *1/2 R² * 1 / (x² + 1/2R²)

E = kQ / (x² + 1/2R²)

Since x >> R, R in the denominator is negligible.

E = kQ / x²

This above equation shows that electric field drops as 1/x² as x approaches .