2. The distance between tow vertical plates in a vacuum is 6.00 cm. A potential

Question

2. The distance between tow vertical plates in a vacuum is 6.00 cm. A potential difference of 300 V is applied between the plates. Point A is located 1.00 cm from the positive plate, point B is at 3.00 cm from it, and point C is at 5.00 cm from it.
a) What is the strength of the electric field at points A, B, and C? Id the electric field constant between parallel plates?
b) What is the potential difference between the positive plate and the points A, B, and C? (Use V = Ed)
c) A positive ion whith a charge of +1.60 x 10^-19 C leaves the positive plate and travels to the negative one. What is the potential energy at the poistive plate? At A? At B? At C? At the negative plate?
Please show all work. Thank you.

Explanation / Answer

A .

V = Ed ;

300 = E . ( 6 e -2 ) ;

E = 5000 N / C ;

yes, the field is constant between the plates ,

B .

+ve plate and A ,

V = 5000 * ( 1 e - 2 ) = 50 V ;

between plate and B ,

V = 5000 * ( 3 e -2 ) = 150 V

between plate and C ;

V = 5000 ( 5 e -2 ) = 250 V

C .

at positive plate : PE = qV = 1.6 e -19 * 300 = 4.8 e -17 J

at A , PE = ( 1.6 e -19 ) ( 300 - 50 ) = 4e -17 J ;

at B , PE = ( 1.6 e -19 ) ( 300 - 150 ) = 2.4 e -17 J ;

at C , PE = ( 1.6 e -19 ) ( 300 - 250) = 8 e -18 J ;

at negative plate , PE = ( 1.6 e -19 ) ( 300 - 300) = 0 J ;

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.