Answer:
The electric field at origin is 3600 N/C
Solution:
As per the question:
Charge density of rod 1, 
Charge density of rod 2, 
Now,
To calculate the electric field at origin:
We know that the electric field due to a long rod is given by:
Also,
(1)
where
K = electrostatic constant = 
R = Distance
= linear charge density
Now,
In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.
At x = - 1 cm = - 0.01 m:
Using eqn (1):
(towards)
Now, at x = 1 cm = 0.01 m :
Using eqn (1):
(towards)
Now, the total field at the origin is the sum of both the fields:
Answer:
The nucleus of an atom contains the majority of the atom’s mass, and is composed of protons and neutrons, which are collectively referred to as nucleons. The much-lighter electrons orbit their atom’s nucleus. The Protons. Protons are positively charged particles found in an atom’s nucleus.
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Answer:
charge and distance
Explanation:
The electric force between the two particles are calculated using the formular:
F = kQ₁Q₂ / d²
where:
F = force.
k= Coulomb's law constant.
Q1 and Q2 are the charges.
d= distance.
the equation above is called Coulomb's law.
It can be seen from the equation above that the electric forces between the objects are majorly affected by the substance's charges and distance.
so the correct option is charge and distance.
Answer:
D.)
Explanation:
the current separates on each branch according to the resistance it experience.
Answer:
A) 0.44 rad/s
B) 3.091 rad/s
Explanation:
Initial angular momentum must be equal to final angular momentum.
W1 = 0.80 rad/s
I = 1360 kgm^2
Initial moment Iw = 0.8 x 1360 = 1088 rad-kg-m^2/s
For second case
I = (65 x 4)4.2 + 1360 = 2452 kg-m2-rad/s
New momentum = 2452 x W2
Where w2 = final angular momentum
= 2452W2
Equating both moment,
1088 = 2452W2
W2 = 0.44 rad/s0
If they were on it before,
Initial momentum
I = 1360 - (60 x 4)4.2
= 352 rad-kg-m^2/s
352W1 = 1088
W1 = 3.091 rad/s
