Oppositely charged parallel plates are separated by 5.33 mm. A potential differe

Question

Oppositely charged parallel plates are separated by 5.33 mm. A potential difference of 600 V existsbetween the plates.

(a) What is the magnitude of the electric fieldbetween the plates?
1      Answer in N/C.

(b) What is the magnitude of the force on an electron between theplates?
2     Answer in N.

(c) How much work must be done on the electron to move it to thenegative plate if it is initially positioned 2.88 mm from the positive plate?
3     Answer in J. (a) What is the magnitude of the electric fieldbetween the plates?
1      Answer in N/C.

(b) What is the magnitude of the force on an electron between theplates?
2     Answer in N.

(c) How much work must be done on the electron to move it to thenegative plate if it is initially positioned 2.88 mm from the positive plate?
3     Answer in J.

Explanation / Answer

Given : .           Potential difference (V) = 600 V .           Distance between the plates (r) = 5.33 mm = 5.33 x 10-3 m . (a)    Electric field is : E  = V / r   = 600 / 5.33 x 10-3 m   = 1.126 x 105   N / C . (b)   Force F = E * q = 1.126 x 105   N / C * 1.602 x 10-19 C   = 1.80 x 10-14 N . (c)   Work W = F *S  =   1.80 x 10-14 N   * 2.88 x 10-3   m   = 5.184 x 10-17   J . Hope this helps u!

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