(a) Determine the electric field strength at a point 1.00 cm to the left of the
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(a) Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in the figure below. (Enter the magnitude of the electric field only.) N/C
6. -3 points SerCP11 15.P.018. My Notes Ask Your (a) Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in the figure below. (Enter the magnitude of the electric field only.) N/C 6.00 pC 1.50,c -2.00,c 3.00 cm- 2.00 cm (b) If a charge of -4.81 uC is placed at this point, what are the magnitude and direction of the force on it? direction Select- Need Help?Read ItExplanation / Answer
The magnitude of the electric field at a distance r from a point particle with charge q is given by:
|E| = |q|*k/r^2
where k is Coulomb's constant = 8.988*10^9 N*m^2/C^2
For a positive charge, the positive direction of the E-field vectors is radially *away* from the charge. For a negative charge, the positive direction is radially *toward* the charge.
The electric field due to multiple point charges is simply the vector sum of the fields due to the individual charges.
For this problem, let the origin be at the point 1cm to the left of the middle particle, and let the x-axis be along the line of particles. Because we are only interested in the value of the field along the x-axis in this problem, the vector sum we have to calculate is fairly simple, and only involves the x-component of the field vector.
With the corrdinate system defined as above, we have that there is a:
+6 microC charge at x = -2
+1.5 microC charge at x = +1
-2 microC charge at x = +3
These give rise to the following fields at x = 0:
E1 = +1 * (6*10^-6 C * 8.988*10^9 N*m^2/C^2)/(-2 cm)^2 = +1.348*10^8 V/m
E2 = -1 * (1.5*10^-6 C * 8.988*10^9 N*m^2/C^2)/(1 cm)^2 = -1.348*10^8 V/m
E3 = +1 * (2*10^-6 C * 8.988*10^9 N*m^2/C^2)/(3 cm)^2 = +1.997*10^7 V/m
Note that 1 N/C = 1 V/m (volt per meter)
Adding these together gives a value of +1.988*10^7 V/m at the origin.
The fields due to the first two particles (the two leftmost ones) exactly cancel each other at the origin. This makes sense because the charge on particle 1 is 4 times that of particle 2, but particle 1 is twice as far from the origin as particle 2. The electric field decays with an inverse square law with distance, so the effect of particle 1 is 1/4 that of particle 2 at the origin.
The force on a charged particle is given by q*E (where E is a vector). If we place a test particle with charge -2 microC at the origin, the force on that particle will be:
(-4.8*10^-6 C) * 1.997*10^7 V/m) = -95.856 N
The force (a vector quantity) is in the negative x direction. This makes sense because we already saw that the fields (and hence the forces) due to the two positive charges cancel one another at the origin, leaving the field due to the rightmost (negatively charged) particle. If we introduce another negatively charged test particle somewhere to the left of the original negatively charged particle, those particles will repel one another; i.e., the test particle will move to the left (the negative x direction).
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