A negatively charged rod of finite length carries charge with a uniform charge per unit length. Sketch the electric field lines
in a plane containing the rod.
1 answer:
Answer:
In the case of a negatively charged rod, the field lines points radially towards the rod.
Explanation:
Field lines pointing towards the negatively charged rod are used to represent the nature of an electric field. Field lines always points in a direction, i.e If the charge is positive, field lines points radially away from the rod and if the charge is negative the field lines points radially towards the rod. So in the case of a negatively charged rod, the field lines points radially towards the rod.
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Answer:
a) v_{p} = 2.83 m / s , b) 50.5º north east
Explanation:
This is a vector problem.
The speed of the ball with respect to the ground is the speed of the ball with respect to Mia plus the speed of Mia with respect to the ground
To make the sum we decompose the speed of the ball in its components
The angle of 30 east of the south, measured from the positive side of the x axis is
θ = 30 + 270 = 300
=
cos 300
= v_{b} sin. 300
v_{bx} = 3.60 cos 300 = 1.8 m / s
v_{by} = 3.60 sin 300 = -3,118 m / s
Let's add speeds on each axis
X axis
vₓ = v_{bx}
vₓ = 1.8 m / s
Y Axis
= v1 - vpy
v_{y} = 5.30 - 3.118
v_{y} = 2.182 m / s
The magnitude of the velocity can be found using the Pythagorean theorem
= √ (vₓ² + v_{y}²)
v_{p} = √ (1.8² + 2.182²)
v_{p} = 2,829 m / s
v_{p} = 2.83 m / s
b) for direction use trigonometry
tan θ = / vₓ
θ = tan ⁺¹ v_{y} / vₓ
θ = tan⁻¹ 2.182 / 1.8
Tea = 50.48º
This address is 50.5º north east
Answer:
1/9 of that just outside the smaller sphere
Explanation:
The electric field strength produced by a charged sphere outside the sphere itself is equal to that produced by a single point charge:
where
k is the Coulomb's constant
Q is the charge on the sphere
r is the distance from the centre of the sphere
Calling R the radius of the first sphere, the electric field just outide the surface of the first sphere is
The second sphere has a radius which is 3 times that of the smaller sphere:
So, the electric field just outside the second sphere is
So, the correct answer is 1/9.