In the diagram, q1 = -6.39*10^-9 C and q2 = +3.22*10^-9 C. What is the electric field at point P? pls help
Answers
Answer:
Below
Explanation:
First draw the vectors that represent both electric fields.
E1 is the elictric field created by q1, E2 is the one created by q2.
● q1 is negative so E1 will point from P.
● q2 is positive so E2 will point out of P
(Picture below)
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The resulting electric field is equal to the sum of the two fields since both vectors are colinear.
Let E be the total field.
● E = E1 + E2
The formula of the electric field intensity is:
● E = K ×(q/d^2)
-K is Coulomb's constant
-d is the distance between the charge and the object ( here P)
-q is the charge
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● E1 = K × (q1/d1^2)
The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)
Coulombs constant is 9×10^9 m^2/C^2
● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]
● E1 = -359.43 N/C
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● E2 = K ×(q2/d^2)
The distance between q2 and P is 0.25 m.
● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]
● E2 = 463.68 N/C
■■■■■■■■■■■■■■■■■■■■■■■■■■
● E = E1 + E2
● E = -359.43+463.68
● E = 105.25 N/C
Related Questions
What did Thomson’s and Rutherford’s experiments have in common? They both used charged particles in their experiments. They both used emission spectra to discover electron clouds. They both compared the atom to an indivisible sphere. They both discovered electron clouds.
Answers
Answer:
Both Thomson and Rutherford used charged particles in their experiments.
Explanation:
Find the x-component of this vector: 47.3 m 39.4 ° Remember, angles are measured from the + x axis. x-component (m)
Answers
The x-component of the vector is 36.55 m.
Components of vectorsThe components of a vector include both x and y components. The x-component is determined along x-axis, while y-component is determined along y-axis.
Vx = Vcosθ
Vy = Vsinθ
where;
Vx is the x-component of the vectorVy is the y-component of the vectorX-components of the vectorThe x-component of the vector is calculated as follows;
x = 47.3 x cos(39.4)
x = 36.55 m
Thus, the x-component of the vector is 36.55 m.
Learn more about x-component of vector here: https://brainly.com/question/13416288
What is the average velocity of atoms in 2.00 mol of neon (a monatomic gas)
at 308 K? Use the equation: -mv2
2
For m, use 0.02000 kg. Remember that R = 8.31 J/(mol-K).
= ER
3
2
nRT
A. 1540 m/s
B. 876 m/s
C. 87.6 m/s
O D. 15,400 m/s
Answers
Answer:
v = 876 m/s
Explanation:
It is given that,
Number of mol of Neon is 2 mol
Temperature, T = 308 K
Mass, m = 0.02 kg
Value of R - 8.31 J/mol-K
We need to find the average velocity of atoms in 2.00 mol of neon. Neon is a monoatomic gas. Let v is the velocity. So,
[tex]\dfrac{1}{2}mv^2=\dfrac{3}{2}nRT\\\\v=\sqrt{\dfrac{3nRT}{m}} \\\\v=\sqrt{\dfrac{3\times 2\times 8.314\times 308}{0.02}} \\\\v=876.47\ m/s[/tex]
So, the correct option is (B).
fill in the term to complete each sentence
a wave carries___ from one place to another.
mechanical waves carry energy through___.
Answers
Energy
A material medium
A wave carries energy from one place to another. Mechanical waves carry energy through a material medium.
What is a wave?A wave is a phenomenon that flows across a material medium without leaving any lasting mark.
Mechanical or electromagnetic waves can be used to classify them. Transverse and longitudinal are the two main forms.
A wave carries energy from one place to another. Mechanical waves carry energy through a material medium.
Hence the correct options for the fill-in-the-blanks are energy and a material medium.
To learn more about the wave refer to the link;
brainly.com/question/3004869
