A uranium and iron atom reside a distance R = 23.70 nm apart. The uranium atom i

Question

A uranium and iron atom reside a distance R = 23.70 nm apart. The uranium atom is singly ionized; the iron atom is doubly ionized. Calculate the distance r from the uranium atom necessary for an electron to reside in equilibrium. Ignore the insignificant gravitational attraction between the particles.

An electron is located on the x axis at x0 = -8.99 × 10-6 m. Find the magnitude and direction of the electric field at x = 1.13 × 10-6 m on the x axis due to this electron. Magnitude:  and Direction:

A small 1.25-g plastic ball that has a charge q of 1.35 C is suspended by a string that has a length L of 1.00 m in a uniform electric field, as shown in the figure. If the ball is in equilibrium when the string makes a 9.80° angle with the vertical as indicated by , what is the electric field strength E?

Two charges are located in the x–y plane. If q1 = -2.75 nC and is located at x = 0.00 m, y = 0.680 m and the second charge has magnitude of q2 = 4.40 nC and is located at x = 1.10 m, y = 0.650 m, calculate the x and y components, Ex and Ey, of the electric field, , in component form at the origin, (0,0). The Coulomb Force constant is 1/(4 0) = 8.99 × 109 N·m2/C2.

Explanation / Answer

(1)

Given that,

Distance b/w  uranium and iron atom, R = 23.70 nm

Lets consider an electron is placed b/w  uranium and iron atom.

Distance b/w electron and uranium = r

Distance b/w electron and iron atom = 23.7 nm - r

Charge on an electron, q1 = 1.6*10^(-19) C

Charge on uranium U+(singly charged), q2 = 1.6*10^(-19) C

Charge on iron (Fe+2), q3 = 2*1.6*10^(-19) C

From the equilibrium of electric force,

kq1q2 / r1^2 = kq1q3 / r2^2

(1.6*10^(-19))^2 / r^2 = 2 * (1.6*10^(-19))^2 / (23.7 - r)^2

1 / r^2 = 2 / (23.7 - r)^2

r = 9.81 nm

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