1-A small sphere of charge q1 = 0.792 µC hangs from the end of a spring as in fi

Question

1-A small sphere of charge q1 = 0.792 µC hangs from the end of a spring as in figure (a). When another small sphere of charge q2 = 0.576 µC is held beneath the first sphere as in figure (b), the spring stretches by d = 3.21 cm from its original length and reaches a new equilibrium position with a separation between the charges of r = 4.60 cm. What is the force constant of the spring?

2-Two point charges lie along the y-axis. A charge of q1 = -6 C is at y = 6.0 m, and a charge of q2 = -7.5 C is at y = -4.0 m. Locate the point (other than infinity) at which the total electric field is zero.?

Explanation / Answer

1)

The additional force (F) on the spring is due to electrical attraction ..
F = kq1.q2/r²
F = 9.0^9 x 0.792^-6C x 0.576^-6C / (0.046m)² = 1.94 N

Force constant = F/d = 1.94N / 0.06m .. .. k = 32.33 N/m

2)

Since the charges are both negative, they will produce equal and opposite components along the "y" axis at some point between the two charges. The E component due to the upper charge (Q1) is:

E1 = Q1/(4*pi**r^2) = Q1/(4*pi**(6 - y)^2)

The E component due to the lower charge is:

E2 = -Q2/(4*pi**r^2) = -Q2/(4*pi**(y+4)^2)

The negative sign handles the fact that the E-field from Q2 points toward the "-y" direction for "y" values greater than -4. At the point where E1+E2 = 0:

Q1/(4*pi**(6-y)^2) = Q2/(4*pi**(y+4)^2)

Notice that the 4*pi* cancels, and that you can take the square root of both sides:

sqrt(|Q1|)/(6 - y) = sqrt(|Q2|)/(y+4)

Cross multiply:

sqrt(|Q1|)*(y+4) = sqrt(|Q2|)*(6-y)

Collect terms:

(sqrt(|Q1|) + sqrt(|Q2|))*y = sqrt(|Q2|)*6 - sqrt(|Q1|)*4

y = ( 6*sqrt(|Q2|) - 4*sqrt(|Q1|))/(sqrt(|Q1|) + sqrt(|Q2|))

y = 1.1788m

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

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