A particle of charge +13 µC and mass 3.57 10-5 kg is released from rest in a region where there is a constant electric field of +495 N/C. What is the displacement of the particle after a time of 1.50 10-2 s?

20.3 mm

Explanation:

The force  on the charge F = qE where q = charge = +13 µC = +13 × 10⁻⁶ C and E = electric field intensity = +495 N/C.

The acceleration of the charge is thus a = F/m where m = mass of charge = 3.57 × 10⁻⁵ kg.

Its displacement is gotten from

s = ut + at²/2  where u = initial speed = 0 (since the charge starts from rest) and t = 1.5 × 10⁻² s.

s = 0 +at²/2 = at²/2

Substituting a and t into s we have

s = qEt²/2m = +13 × 10⁻⁶ C × +495 N/C × (1.5 × 10⁻² s)²/(2 × 3.57 × 10⁻⁵ kg) = 0.0203 m = 20.3 mm

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2