(a) The drum of a photocopying machine has a length of 42 cm and a diameter of 12 cm. The electric field just above the surface of the charged drum has a magnitude E of 3.0 × 105 N/C. What is the total charge on the drum? (b) The manufacturer wishes to produce a desktop version of the machine. This requires reducing the drum length to 22 cm and the diameter to 6.2 cm. The electric field at the drum surface must not change. What must be the charge on this new drum?

Step-by-step explanation:

The drum is in the form of charged cylinder .

Electric field near a charged cylinder = λ /2πε₀r

λ is linear charge density ,

3 x 10⁵ = λ /(2 x 3.14 x 8.85 x 10⁻¹² x 0.06)

λ = 2 x 3.14 x 8.85 x 10⁻¹² x 0.06  x 3 x 10⁵

= 10 x 10⁻⁷ C/m

Total charge on the cylinder

= .42 x 10 x 10⁻⁷

= 4.2 x  10⁻⁷ C

b )

New diameter = 6.2 cm

radius = .031 m , for same electric field

3 x 10⁵ = λ /(2 x 3.14 x .031 x 8.85 x 10⁻¹² )

λ = 2 x 3.14 x .031 x 8.85 x 10⁻¹² x 3 x 10⁵

= 5.168 x 10⁻⁷

Charge on the cylinder

=  5.168 x 10⁻⁷ x .22

= 1.25 x 10⁻⁷ C

Charge on new drum

= 1.25 x 10⁻⁷ C

answer: the first one

step-by-step explanation: use a calculator and divide the minutes by the dollars.

it will be a cylinder. with a diameter of 8 and 4 single pieces high

step-by-step explanation: