Question

Two point charges lie on the x axis. A charge of 6.4 μC is at the origin, and a charge of -9.5 μC is at x=10.0cm.

a) What is the net electric field at x=−4.0cm? (N/C?)

b)What is the net electric field at x=+4.0cm? (N/C?)

Answer 1

Answer:

a)  $Ep=3.16*10^{7} (-i) \frac{N}{m}$  ,at x = −4.0cm

b)$Ep=5.97*10^{7} (+i) \frac{N}{m}$,  at x=10.0cm

Explanation:

To solve this problem we apply the theory of the magnetic field:

Formula to calculate the magnetic field (E) at a point P due to an electric charge(q):

$E=\frac{k*q}{d^{2} }$    Formula (1)

K: Coulomb constant in $\frac{N}{m^{2}*C^{2}  }$

q= Electric charge in Coulombs( C)

d= Distance from location of q to point P in meters (m)

Concepts that we must take into account

:

• The total electric field due to a group of charges is equal to the vector sum of the electric fields of all charges.
• The electric field lines leave a positive charge and enter a negative charge.

1μC=$10^{-6} C$

1cm=$10^{-2} m$

Problem development

Case a) calculation of the net electric field at x = −4.0cm

Data:

q1=6.4μC =$6.4 *10^{-6} C$, q2=-9.5μC =$-9.5 *10^{-6} C$

$d1=4cm=4*10^{-2}m$,   $d2=14cm=14*10^{-2}m$

$k=8.99*10^{9} \frac{N}{m^{2}*C^{2}  }$

Ep=E1+E2 : Total electric field at point P

We use formula 1 to calculate the electric field:

Ep=$\frac{k*q1}{d1^{2} }$ + $\frac{k*q2}{d2^{2} }$ : If we factor K and replace the data, then:

$Ep=8.99*10^{9} (\frac{6.4*10^{-6} }{(4*10^{-2})^{2} }(-i) +\frac{9.5*10^{-6}}{(14*10^{-2})^{2}} (+i))$:if we use factor  $(\frac{10^{-6} }{10^{-4} } )$,then:

$Ep: 8.99*10^{9} (\frac{10^{-6} }{10^{-4} } ) (\frac{6.4}{16} (-i) +(\frac{9.5}{196}(i))$

$Ep=(-3.6+0.4357)*10^{7} (i) \frac{N}{C}$

$Ep=3.16*10^{7} (-i) \frac{N}{m}$

Case b) calculation of the net electric field at x = +4.0cm

Data:

q1=6.4μC =$6.4 *10^{-6} C$, q2=-9.5μC =$-9.5 *10^{-6} C$

$d1=4cm=4*10^{-2}m$,   $d2=6cm=6*10^{-2}m$

$k=8.99*10^{9} \frac{N}{m^{2}*C^{2}  }$

Ep=E1+E2 : Total electric field at point P

We use formula 1 to calculate the electric field:

Ep=$\frac{k*q1}{d1^{2} }$ + $\frac{k*q2}{d2^{2} }$ : If we factor K and replace the data, then:

$Ep=8.99*10^{9} (\frac{6.4*10^{-6} }{(4*10^{-2})^{2} }(+i) +\frac{9.5*10^{-6}}{(6*10^{-2})^{2}} (+i))$: If we factor $(\frac{10^{-6} }{10^{-4} } )$,then:

$Ep: 8.99*10^{9} (\frac{10^{-6} }{10^{-4} } ) (\frac{6.4}{16} (+i) +(\frac{9.5}{36}(i))$

$Ep=(+3.6+2.37)*10^{7} (i) \frac{N}{C}$

$Ep=5.97*10^{7} (+i) \frac{N}{m}$