**Problem Overview: Calculating Net Force Using Coulomb’s Law**Two charges, +8 μC and +10 μC, are placed along the x-axis. The first charge is at the origin (\(x = 0\)) and the second charge at \(x = +1 \, \text{m}\). Determine the magnitude and direction of the net force on a -3 nC charge when placed at the following locations.**Steps for Solution:**1. **Use of Coulomb's Law:** - Review Section 5.3 Coulomb’s Law. - Calculate the force due to each of the two charges. - Sum the forces for the net force, considering directions for accuracy. Part (c) will require a vector sum.2. **Problem Locations:** - **a.** Halfway between the two charges: - Determine the magnitude and select the direction from dropdown options. - **b.** On the x-axis at \(x = -0.5 \, \text{m}\): - Determine the magnitude and select the direction from dropdown options. - **c.** At the coordinate (1 m, 0.5 m): - This requires considering a direction perpendicular to the line joining the two fixed charges.**Hints and Calculations for Part (c):**- This is a 2-dimensional problem.- The charge at \(x = 1 \, \text{m}\) affects force along the y-direction.- Charge at the origin affects force along the line joining \((x, y) = (0, 0)\) and \((x, y) = (1 \, \text{m}, 0.5 \, \text{m})\).- Express each force vector in terms of \(x\) and \(y\) components, sum them, and calculate the resultant magnitude using the Pythagorean theorem.- Use \(\tan(\theta) = \frac{F_y}{F_x}\) to find direction.**Answer Submission:**- Enter the magnitude of force in mN and direction in degrees.- Use the "Submit Question" option for checking answers.**Question Help:**- Access support through "Message Instructor" for clarifications.For detailed resources and formulas, refer to the mentioned section on Coulomb’s Law.

Given,The magnitude of charge 1, q1 = 8 μC at originThe magnitude of charge 2, q2 = 10 μC at x = +1 mThe magnitude and diretion of net force on -3 nC (q3),a) Halfway between the two charges,Force on charge 3, F→3 =F→13+F→23r13=r23 =r=0.5 mF→13=-14πε08 μC×3 nCr2i^=-14πε08 μC×3 nC(0.5)2i^=-9×24×10-60.25i^F→13=-0.000864 i^ N=-0.864 mN i^F→23=14πε010 μC×3 nCr2i^=9×30×10-60.25i^=1.08 mN i^F→3 =F→13+F→23=1.08 mN i^-0.864 mN i^ = 0.216 mN i^Force on charge 3, halfway between the two charges is 0.216 mN i^or 0.216 mN along positive x axis.b) charge 3 at x =-0.5 m,Force on charge 3, F→3 =F→13+F→23r13=0.5 mr23=1.5 mF→13=14πε08 μC×3 nCr132i^=14πε08 μC×3 nC(0.5)2i^=9×24×10-60.25i^F→13=0.864 mN i^F→23=14πε010 μC×3 nCr232i^=14πε010 μC×3 nC(1.5)2i^=9×30×10-62.25i^F→23=0.12 mN i^F→3=F→13+F→23=0.864 mN i^+0.12 mN i^= 0.984 mN i^Force on charge 3 is 0.984 mN i^Force on charge 3 is 0.984 mN along positive x axisc) charge 3 at (x,y) =(1m, 0.5 m)Force on charge 3, F→3 =F→13+F→23r13=1 m i^+ 0.5 m j^r23=0.5 m j^F→13= F→13i^+ F→13j^=-14πε08 μC×3 nC(1)2i^-14πε08 μC×3 nC(0.5)2 j^F→13=-0.216 mN i^-0.864 mN j^F→23=-14πε010 μC×3 nC(0.5)2 j^=-1.08 mN j^F→3 =F→13+F→23=-0.216 mN i^-0.864 mN j^-1.08 mN j^F→3=-0.216 mN i^-1.944 mNj^Magnitude of force is (-0.216)2+(-1.944)2Magnitude of force is 1.956 mNDirection is θ=tan-1-1.944-0.216 =83.66°Magnitude of force is 1.956 mN and direction is 83.66° from - x axis inCCW direction

Answered: Two charges +8 μC and +10 μC are placed…

Two charges +8 μC and +10 μC are placed along the x axis, with the first charge at the origin (0) and the second charge at x = +1 m. Find the magnitude and direction of the net force on a -3 nC charge when placed at the following locations below. -