Answer:
Vb = k Q / r r <R
Vb = k q / R³ (R² - r²) r >R
Explanation:
The electic potential is defined by
ΔV = - ∫ E .ds
We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product
VB - VA = - ∫ E dr
Let's substitute every equation they give us and we find out
r> R
Va = - ∫ (k Q / r²) dr
-Va = - k Q (- 1 / r)
We evaluate with it Va = 0 for r = infinity
Vb = k Q / r r <R
We perform the calculation of the power with the expression of the electric field that they give us
Vb = - int (kQ / R3 r) dr
We integrate and evaluate from the starting point r = R to the final point r <R
Vb = ∫kq / R³ r dr
Vb = k q / R³ (R² - r²)
This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity
Answer:
P = 1 x 10⁸ Pa
Explanation:
given,
radius = 2.0 ×10⁻¹⁰ m
Temperature
T = 300 K
Volume of gas molecule =


V = 33.51 x 10⁻³⁰ m³
we know,
P V = 1 . k T
k = 1.38 x 10⁻²³ J/K
P(33.51 x 10⁻³⁰) = 1 . (1.38 x 10⁻²³) x 300
P = 1.235 x 10⁸ Pa
for 1 significant figure
P = 1 x 10⁸ Pa
Answer:
mass of the object is 2.18 kg
Explanation:
Given
Force (F) = 8.5 N = 8.5 kg.m/
acceleration (a) = 3.9 m/
Mass (m) = ?
We know that the newton's second law of motion gives the relation between mass of ab object. force acted upon and the amount the object is accelerated. It is expressed in the form of an equation:
F = ma
mass, m = F/a
= 
= 2.18 kg