Coulomb's Law — Question 10

Section 21-3 · Physics pages 3430–3431

Question 10 Balancing one corner of a square does not balance the whole configuration A square of side a has positive charge q at (0, 0), (a, 0) and (0, a). An adjustable charge Q is fixed at (a, a). Use Coulomb law with constant k and ignore gravity. q q q Q a Schematic; labels, not drawing lengths, specify the data. (a) Find Q so the electrostatic force on the origin charge is zero. (b) For that value of Q, find the force components on the charge at (a, 0) in units F0 = kq2/a2. Determine whether all four charges can remain at rest without supports.
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Question 10 - Solution Balancing one corner of a square does not balance the whole configuration q q q Q a (a) Method and calculation Fadjacent,on origin = −F0(ˆi +ˆj). FQ,on origin = −kqQ 2 √ 2a2 (ˆi +ˆj). q2 + qQ/(2 √ 2) = 0 ⇒Q = −2 √ 2q. (b) Interpretation and further calculation Fx(a, 0) F0 = 1 + 1 2 √ 2, Fy(a, 0) F0 = 2 √ 2 − 1 2 √ 2. F(a, 0)̸ = 0. The origin charge is balanced, but at least one other charge is not. Fixed source positions therefore require external support forces. Physical check The diagonal separation is √ 2a and its force must also be projected by 1/ √ 2 onto each axis.

Question 10

Balancing one corner of a square does not balance the whole configuration

A square of side aa has positive charge qq at (0,0)(0,0), (a,0)(a,0) and (0,a)(0,a). An adjustable charge QQ is fixed at (a,a)(a,a). Use Coulomb law with constant kk and ignore gravity.

Diagram shown in the original SVG worksheet.

  1. Find QQ so the electrostatic force on the origin charge is zero.

  2. For that value of QQ, find the force components on the charge at (a,0)(a,0) in units F0=kq2/a2F_0=kq^2/a^2. Determine whether all four charges can remain at rest without supports.

Question 10 - Solution

Balancing one corner of a square does not balance the whole configuration

Diagram shown in the original SVG worksheet.

(a) Method and calculation

𝑭adjacent,onorigin=−F0(𝒊̂+𝒋̂).\mathbf F_{\mathrm{adjacent,on\ origin}}=-F_0(\hat{\mathbf i}+\hat{\mathbf j}).𝑭Q,onorigin=−kqQ22a2(𝒊̂+𝒋̂).\mathbf F_{Q,\mathrm{on\ origin}}=-\frac{kqQ}{2\sqrt2a^2}(\hat{\mathbf i}+\hat{\mathbf j}).q2+qQ/(22)=0⇒Q=−22q.q^2+qQ/(2\sqrt2)=0\Rightarrow Q=-2\sqrt2q.

(b) Interpretation and further calculation

Fx(a,0)F0=1+122,Fy(a,0)F0=22−122.\frac{F_x(a,0)}{F_0}=1+\frac1{2\sqrt2},\quad\frac{F_y(a,0)}{F_0}=2\sqrt2-\frac1{2\sqrt2}.𝑭(a,0)≠𝟎.\mathbf F(a,0)\ne\mathbf0.The origin charge is balanced, but at least one other charge is not. Fixed source positions therefore require external support forces.

Physical check

The diagonal separation is 2a\sqrt2a and its force must also be projected by 1/21/\sqrt2 onto each axis.