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Electric Field of Equilateral Triangle Charges

Created by @nathanedwards
 at October 31st 2023, 7:26:33 pm.
AP_Physics_2-Questions
#electric-field
#ap-physics-2
#equilateral-triangle

AP Physics 2 Exam Question:

Consider three point charges arranged in an equilateral triangle configuration with side length a, as shown below.

  Q1
   •
     \
     a \
        \ 
 _______\_________
    Q2     Q3

a) Derive an expression for the net electric field at the center of the triangle due to the charges Q1, Q2, and Q3, in terms of their magnitudes and a. Show all your steps.

b) If Q1 = Q, Q2 = 2Q, and Q3 = Q, calculate the magnitude and direction of the net electric field at the center of the triangle when a = 0.1 m. Assume that k = 8.99 × 10^9 N m^2/C^2.

Solution:

a) To find the net electric field at the center due to the charges Q1, Q2, and Q3, we need to consider the electric fields contributed by each charge individually and then sum them up vectorially.

Let's assume the center of the triangle as point O, located at a perpendicular distance h from the line connecting Q2 and Q3. We will calculate the electric fields at the center due to each individual charge using Coulomb's law.

The electric field due to Q1 at point O is given by:

E1=k⋅Q1r12 E_1 = \frac{{k \cdot Q_1}}{{r_1^2}}

Since the distance between O and Q1 is a/2, we can substitute r1 = a/2 in the above equation.

Next, we find the electric field due to Q2. Notice that Q2 lies on the line connecting Q1 and O, so the electric field due to Q2 at point O is zero. This is because the electric field vectors at O and Q2, being equidistant from Q2, have equal magnitudes but opposite directions, hence they cancel out.

Similarly, the electric field due to Q3 at point O is zero.

Therefore, the net electric field at the center O, due to the configuration of charges, is given by:

Enet=E1+E2+E3 E_\text{{net}} = E_1 + E_2 + E_3

Substituting the individual electric fields derived above, we get:

Enet=k⋅Q1(a/2)2 E_\text{{net}} = \frac{{k \cdot Q_1}}{{(a/2)^2}}

Simplifying,

Enet=4⋅k⋅Q1a2 E_\text{{net}} = \frac{{4 \cdot k \cdot Q_1}}{{a^2}}

Hence, the expression for the net electric field at the center due to the charges Q1, Q2, and Q3 is 4⋅k⋅Q1a2\frac{{4 \cdot k \cdot Q_1}}{{a^2}}.

b) Given that Q1 = Q, Q2 = 2Q, Q3 = Q, and a = 0.1 m, we can substitute these values into the expression for the net electric field at the center derived in part a.

Thus,

Enet=4⋅k⋅Q1a2 E_\text{{net}} = \frac{{4 \cdot k \cdot Q_1}}{{a^2}}

Plugging in the values,

E_\text{{net}} = \frac{{4 \cdot (8.99 \times 10^9 \, \text{{N m}}^2/\text{{C}}^2) \cdot Q}}}{{(0.1 \, \text{{m}})^2}}

Simplifying,

Enet=35960⋅Q N/C E_\text{{net}} = 35960 \cdot Q \, \text{{N/C}}

Therefore, the magnitude of the net electric field at the center of the triangle is 35960⋅Q N/C35960 \cdot Q \, \text{{N/C}}.

Since all the charges Q1, Q2, and Q3 are positive, their electric fields at the center will have the same direction. According to the arrangement of charges, the net electric field will be directed radially outward from the center of the triangle.

Hence, the direction of the net electric field at the center of the triangle is outward radially away from the center.

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