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

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

Answers

Answer 1

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)

■■■■■■■■■■■■■■■■■■■■■■■■■■

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

■■■■■■■■■■■■■■■■■■■■■■■■■■

● 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

■■■■■■■■■■■■■■■■■■■■■■■■■■

● 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

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

Related Questions

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 vectors

The 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 vector

X-components of the vector

The 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.

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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
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Answers

Answer:

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Explanation:

It is given that,

Number of mol of Neon is 2 mol

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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,

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I think it is 887m/s I hope this helps if not I’m really sry

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a wave carries___ from one place to another.
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Answers

Energy

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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.

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brainly.com/question/3004869

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Answers

Answer:

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Explanation:

Theo both used charged particles
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