Please help me in both of them problem A test charge of +3 times 10^-7 C is loca

Question

Please help me in both of them problem A test charge of +3 times 10^-7 C is located 5 cm to the right of a charge of 4 times 10^C and 10 cm to the left of a charge of -1 times 10^6 C. The three charges lie on a straight line. (a) Find the force on the test charge. (b) Find the intensity and the potential of electric field at this point. (e) If the test charge is free, with what acceleration it will start the motion? The mass of the object is 1 kg. Occasionally, people who gain static charge by snuffling their feet on the will get electric shock if they touch another people. Why does it happen?

Explanation / Answer

q1 = -4*10^-6 C

q2 = 3*10^-7 C

q3 = -1*10^-6 C

Force on the test charge due to q1:

F1 = k*q1*q2/r^2

= 9*10^9*(4*10^-6)*(3*10^-7)/0.05^2

= 4.32 N/C directed towards left

Similalrly, force on test charge due to charge on right:

F2 = k*q2*q3/r^2

= 9*10^9*(3*10^-7)*(1*10^-6)/0.1^2

= 0.27 N/C to the right

So, force on the test charge = 4.32 - 0.27 = 4.05 N/C to the left <---answer

b)

Intensity = F/q2 = 4.05/(3*10^-7) = 1.35*10^7 N/C

Potential = k*q1/r1 + k*q3/r2 = 9*10^9*(-4*10^-6/0.05 - 1*10^-6/0.1)

= -8.1*10^-5 V <-------answer

c)

F = m*a

So, a = F/m = 4.05/1 = 4.05 m/s2 <--------answer

6)

When you snuffle your feet on the carpet enough static charges get transferred from the carpet to you.Now, as you touch another person, the excess charge gets a short enough path for its transfer to another body and it you discharge the static charges as an electric shock.

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