Answer:
3.135 kN/C
Explanation:
The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}
Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m
So, E = qz/{4πε₀[√(z² + R²)]³}
E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}
E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}
E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}
E = 6.839 × 10³ Cm²/[1.297 m]³F
E = 6.839 × 10³ Cm²/2.182 m³F
E = 3.135 × 10³ V/m
E = 3.135 × 10³ N/C
E = 3.135 kN/C
Answer: Really
Explanation:
Just look it up for this page and maybe you will find an anwser sheet.
Answer:
Distance covered is equal to all the distance traveled.
So for example, if you go from A to B, and then from B to C, the total distance covered is AB + BC.
Displacement is equal to the difference between the final position and the initial position.
So if we go from A to B, the displacement is simply the line AB.
While if we go from A to B, and then from B to C, the displacement will be a segment that directly connects A and C, such that:
displacement = √( (AB)^2 + (BC)^2)
Now, if we want to find the points such that the magnitude of the distance covered is equal to the magnitude of the displacement, we need to look at the pairs that are directly connected by a straight line.
Those are:
A to B ( or B to A)
B to C (or C to B)
C to D (or D to C)
