Final Answer:
The electric field at the point (0, 0.800 m) is 3.10E5 N/C in the negative x-direction.
Step-by-step explanation:
The electric field at a point due to two or more charges is the vector sum of the electric fields due to individual charges present at that point. The electric field due to a single point charge is given by E = KQ/r2, where K is the Coulomb constant (8.99E9 Nm2/C2), Q is the charge and r is the distance of the point from the charge.
In the given problem, a 2.00 nC point charge is at the origin and a second -5.00 nC point charge is on the x-axis at x = 0.800 m. The electric field at the point (0, 0.800 m) is the vector sum of the electric fields due to the two charges.
The electric field due to the 2.00 nC point charge at the origin is given by E1 = KQ1/r12, where K is the Coulomb constant, Q1 is the charge and r1 is the distance of the point from the charge. Therefore,
E1 = (8.99E9 Nm2/C2) (2.00 nC)/(0.800 m)2 = 3.50E5 N/C.
Similarly, the electric field due to the -5.00 nC point charge at x = 0.800 m is given by E2 = KQ2/r22, where K is the Coulomb constant, Q2 is the charge and r2 is the distance of the point from the charge. Therefore,
E2 = (8.99E9 Nm2/C2) (-5.00 nC)/(0.800 m)2 = -4.00E5 N/C.
The electric field at the point (0, 0.800 m) is the vector sum of the electric fields due to the two charges, E = E1 + E2.
Therefore,
E = (3.50E5 N/C) + (-4.00E5 N/C) = -3.10E5 N/C in the negative x-direction.
Hence, the electric field at the point (0, 0.800 m) is -3.10E5 N/C in the negative x-direction.