A charge 50 µc has electric field strength of 360 NC-1 at a certain point. The electric field strength due to another charge 120 µc kept at the same distance apart and in the same medium is?
864 NC-1
150 NC-1
144 NC-1
18 NC-1
Explanation
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Discussions (27)
This is wrong E = (F)/q
but F=Qq/d(square)
E=kQ/d(square) where k is a constant
so solve for d using the equation above and since d(distance) is kept the same
solve for E and you will get exactly 864 ANSWER IS A.
We know that E = F/q and F = kQq/r²
Where,
E is the electric field intensity
F is the force of attraction between the two charges
k is the constant 9 × 10⁹ Nm²/c²
Q and q are the two charges in question
r is the distance between the two charges
Recall that E = F/q and F = kQq/r²
So, E = kQ/r²
From the given parameters, for the charge Q = 50µc, E = 360N/c
So, 360N/c = (9 × 10⁹ Nm²/c² × 50 × 10^-6 c)/r²
Making r² the subject of formula, we get
r² = (9 × 10⁹ × 50 × 10^-6 )/360
= (450 × 10^3)/360
r² = 1.25 × 10^3
For another charge q = 120µc which is kept at the same distance apart and in the same medium, we are told to find the electric field intensity at this point, thus
E = kq/r², where r² is still = 1.25 × 10^3
= (9 × 10⁹ × 120 × 10^-6 )/1.25 × 10^3
= (1080 × 10^3)/1.25 × 10^3
E = 864N/c, which is option A
Electric Field Strength
The electric field strength (E) due to a point charge is given by:
E = k × q / r²
Given the same distance (r) and medium:
Proportionality:
E ∝ q
Given:
1. q₁ = 50 μC = 50 × 10⁻⁶ C
2. E₁ = 360 N/C
3. q₂ = 120 μC = 120 × 10⁻⁶ C
Find E₂:
E₂ / E₁ = q₂ / q₁
E₂ = E₁ × (q₂ / q₁)
= 360 × (120 × 10⁻⁶) / (50 × 10⁻⁶)
= 360 × 2.4
= 864 N/C
Electric Field Strength due to 120 μC Charge:
E₂ = 864 N/C
Oh. Pls I know it's last minute but Myschool plssssssss 🥹🥹 is it possible to simplify most of your explanations...plss.
Thank you.
Some stuffs you explain make feel like I grew up with Micheal Faraday or
maybe I helped him out with his inventions....😪😪. They're quite complex sometimes that's all, really.
More grease to your elbows
pls can someone carefully solve this and while doing so show how exactly d² was gotten because from the question i couldn't find the distance d and it quite confused my understanding on how you guys got 864
My school, there was no need doing all those unnecessary stuff you did.
two charges placed in the same medium exerts equal and opposite forces.
The first charge would exert a force of 360 × 50 = 18000 µN. [Since E = F/Q]
The other charge would exert exactly the same force of 18000 µN.
Hence, the ekectric field due to the second charge is E = 18000 µN/120 µc = 150 N/C.
E =
F
q
and Q1 = 50μF, E1 = 360μF, E2 = ?, Q2 = 120μF
Using E =
F
q
.
So we have F = Q × E
F = 50 × 10−6 × 360 = 0.018N
When E2 = F/q
0.018
120×10−6
= 150NC
The correct answer is option B
Q1=50uc, E1= 360NC^1, Q2=120NC, E2=??
RECALL THAT E= F/Q, THEN F = EQ
SINCE THEIR DISTANCES ARE DSAME, YOU HAVE TO FIND F FIRST
F= EQ=50*360=18000
NOW THE SECOND PART OF THE QUESTION
F=EQ BUT HERE E IS UNKNOWN
E= F/Q = 18000/120 =150NC^1
