7) A charge of +2 C is at the origin. When charge Q is placed at 2 m along the p

Question

7) A charge of +2 C is at the origin. When charge Q is placed at 2 m along the positive x axis, the electrie field at 2 m along the negative x axis becomes zero. What is the value of 0? 2m +2C 8) An electron (charge -1.6x1019 C) moves on a path perpendicular to the direction of a uniform electric field of strength 3.0 N/C, How much work is done on the electron as it moves 20 cm? a. 1.6x10 20 j b. -4.8x1020 J c. 4.8x10-20 J d. zero 9) A proton (+1.6x100 C) moves 10 cm along the direction of an electric field of strength 4 .0 N/C. The electrical potential difference between the proton's initial and ending points is: a. 4.8x10 19 v b. 0.40 V c. 0.044 V. d. 6.4x10-19 v 10) What is the equivalent capacitance between points a and b? All capacitors are 2.0 uF. -? ?? a. 1.2 uF b. 0.25 uF c. 4.0?F d. 0.60 pF

Explanation / Answer

Q7.

U the formula E = kq / r^2 where
k = 8.99*10^9 (N*m/C^2)
q is the magnitude of the charge
r is the distance between the charge and the point.
Since the field at x=-2 m is zero, the nature of Q must be opposite to the charge at the origin, hence Q is negative.
Now, since the filed at x=-2 m is zero, the sum of the fields due the charge at the origin and Q must be zero.

Thus, k*2/2^2 + k*Q/4^2 = 0
=> 2/2^2 + Q/4^2 = 0
Solving, Q = -8C

So, option C.

Pls post other questions separately

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