# Compute the tension in string due to charges in equilibrium

Three equal charges, Q coulombs are located at the vertices of a triangle as shown in the figure. The charges are connected together by strings as shown in the figure. If the length of each string is L, determine the tension in the strings.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

View the attachment.

Since all the charges are of equal sign, each Q is experiencing a repulsive force from the other two. Since the string holds them all in equilibrium all these forces add up to zero.

I.e., F13 + F12 + T1 + T3 = 0

We can write F13 = Q1 E3 and F12 = Q1 E2 Where E3 and E2 are given by coulomb's law. F13 is the force acting on charge Q1 due to charge Q3 and so on.

E3 = k {Q3 / r32} where k = 1/4Ã°Ã¥0 (Of course, there would be a unit vector (r2) showing the direction of the force, from charge Q3 to charge Q1. We will neglect this vector for the time being (but its very important: remember it's there)

E2 = k {Q2 / r22} Substituting back, in the first equation Q1 E3 + Q1 E2 + T1 + T3 = 0

Or, after giving the unit vectors,

k {Q1Q3/ r32} (r3) + k {Q1Q2/ r22} r2 + T1 + T3 = 0

T1 and T3 are also vector quantities, so we can add them with out violating any dimensional rule

T1 = T1 (- r2) and T3 = T3 (- r3)

Above equation gives k {Q1Q3/ r32} (r3) + k {Q1Q2/ r22} (r2) + T1 (- r2) + T3 (- r3) = 0

Combining terms [k {Q1Q3/ r32}- T3] r3 + [k {Q1Q2/ r22}- T1] r2 = 0

In order that the right side is zero, each term on the left must be zero, thus we get,

T3 = k {Q1Q3/ r32} and T1 = k {Q1Q2/ r22 } also we can write T2 = k Q3Q2 / r12 Now you can substitute the values to find the tension in each string. In our problem we have r1 = r2= r3 = L and Q1=Q2=Q3=

View the attachment (figure and explanation)

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