Acharge of 63.0 nc is located at a distance of 3.40 cm from a charge of -47.0 nc. what are the x- and y-components of the electric field at a point p that is directly above the 63.0 nc charge at a distance of 1.40 cm? point p and the two charges are on the vertices of a right triangle.

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/r²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

r: distance from charge q to point P in meters (m)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Equivalences

1nC= 10⁻⁹ C

1cm= 10⁻² m

Graphic attached

The attached graph shows the field due to the charges:

Ep₁: Total field at point P due to charge q₁. As the charge is positive ,the field leaves the charge.

Ep₂: Total field at point P due to charge q₂. As the charge is negative, the field enters the charge.

Known data

q₁ = 63 nC = 63×10⁻⁹ C

q₂ = -47 nC = -47×10⁻⁹ C

k = 8.99*10⁹ N×m²/C²

d₁ = 1.4cm = 1.4×10⁻² m

d₂ = 3.4cm = 3.4×10⁻² m

Calculation of r and β

Problem development

Ep: Total field at point P due to charges q₁ and q₂.

Ep₁ₓ = 0

Calculation of the electric field components at point P

the answer is a 3.8cm

explanation:

cancer risk hope this