A 3.0-μC positive point charge is located at the origin and a 2.0 μC positive point charge is located at
X = 0.00 m,y = 1.0 m. Find the coordinates of the point where the net electric field strength due to these charges is zero.
A) x = 0.00 m, y = 0.55 m
B) x = 0.00 m, y = 0.67 m
C) x = 0.00 m, y = 1.5 m
D) x = 0.00 m, y = 0.60 m

Question

Quizplus · Accepted Answer

The net electric field strength is zero at the point where the electric field strength due to the 3.0-μC charge is equal and opposite to the electric field strength due to the 2.0 μC charge. The electric field strength due to a point charge is given by:$$E = k\frac{q}{r^2}$$where k is Coulomb's constant, q is the charge, and r is the distance from the charge.The electric field strength due to the 3.0-μC charge at the point (x, y) is:$$E_1 = k\frac{3.0	imes10^{-6}	ext{ C}}{(x^2 + y^2)}$$The electric field strength due to the 2.0 μC charge at the point (x, y) is:$$E_2 = k\frac{2.0	imes10^{-6}	ext{ C}}{[(x-0.00	ext{ m})^2 + (y-1.0	ext{ m})^2]}$$The net electric field strength is zero when:$$E_1 = E_2$$$$k\frac{3.0	imes10^{-6}	ext{ C}}{(x^2 + y^2)} = k\frac{2.0	imes10^{-6}	ext{ C}}{[(x-0.00	ext{ m})^2 + (y-1.0	ext{ m})^2]}$$$$\frac{x^2 + y^2}{(x-0.00	ext{ m})^2 + (y-1.0	ext{ m})^2} = \frac{3}{2}$$$$2x^2 + 2y^2 = 3(x^2 - 2x	imes0.00	ext{ m} + 0.00^2	ext{ m}^2 + y^2 - 2y	imes1.0	ext{ m} + 1.0^2	ext{ m}^2)$$$$2x^2 + 2y^2 = 3x^2 - 6x + 3y^2 - 6y + 3$$$$x^2 + 6x + y^2 + 6y - 3 = 0$$$$(x + 3)^2 + (y + 3)^2 = 12$$The point where the net electric field strength is zero is at the center of the circle defined by the equation:$$(x + 3)^2 + (y + 3)^2 = 12$$The center of the circle is at the point (-3, -3). Therefore, the coordinates of the point where the net electric field strength is zero are:$$x = -3 + 0.00 = 0.00	ext{ m}$$$$y = -3 + 3 = 0.60	ext{ m}$$

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