Calculate the magnitude of the electric field at the origin due to the following distribution of charges:
Calculate the magnitude of the electric field at the origin due to the following distribution of charges: -q at (x,y) = (a,a), -q at (-a,-a), +q at (-a,-a) and +q at (-a,a). Where q = 5.85 × 10-7 C and a = 2.10 cm.
Set up an 8-point compass at the origin, where north points along the positive y-axis, such as that shown in the diagram to the right. What is the direction of the electric field at the origin created by the charge distribution given in the previous part? (for example, `N',`NE', etc., or if no net field enter `NIL')
Homework Answers
Electric field due to the four charges at the origin are
, 
, 
,
Net electric field at origin is 
Direction of electric field is along East (E)
Electric field at the origin is 
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Calculate the magnitude of the electric field at the origin due to the following distribution of charges:
Distribution of Charges Part A Calculate the magnitude of the electric field at the origin due...
Distribution of Charges Part A Calculate the magnitude of the electric field at the origin due to the following distribution of charges: -q at (x,y) - (a,a),-q at (a,-a), +q at (-a,-a) and +q at (-a,a), where q = 3.25 × 10-7 C . and a 4.40 cm2 Submit Answer Tries 0/6 Part B Set up an 8-point compass at the origin, where north points along the positive y-axis, such as that shown in the diagram to the right. What...
Distribution of Charges Calculate the resnitude of the electric field et the ongin due to the...
Distribution of Charges Calculate the resnitude of the electric field et the ongin due to the fulv 'ng distribution of charqes: q dt(x,y: d d -q dt d d ,-q et d djand +q dt d d where q = 3.5ข x 10 7 Cend d =。10 cm Submit Anzwer Trics /10 set up an ¥ pomt oor pass at the rigin, where north pu nts d ony he postive γ axis, such as t at sho n n he...
An electric field can be created by a single charge or a distribution of charges. The...
An electric field can be created by a single charge or a distribution of charges. The electric field a distance r from a point charge q' has magnitude {E}=k\frac{|q'|}{r^2}. The electric field points away from positive charges and toward negative charges. A distribution of charges creates an electric field that can be found by taking the vector sum of the fields created by individual point charges. Note that if a charge q is placed in an electric field created by...
a) Calculate the magnitude of the net electric field at the origin due to these two point charges. b) Calculate...
a) Calculate the magnitude of the net electric field at the origin due to these two point charges. b) Calculate the direction of the net electric field at the origin due to these two point charges.
For starters, calculate the magnitude and direction of the electric field due only to charge q1...
For starters, calculate the
magnitude and direction of the electric field due only to
charge q1 at this point.
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Incompatible units. No conversion found between "N" and the
required units.
Tries 0/10
Previous Tries
Calculate the magnitude and direction of the electric field
due only to charge q2 at this
point.
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Up
Left
Right
Tries 0/10
Calculate the magnitude and direction of the electric field
due only to charge q3 at this
point.
Down
Up...
The diagram below shows a dipole centered at the total electric field at A, when q...
The diagram below shows a dipole centered at the total electric field at A, when q = 5.05×10-7 C and L-49.2 cm. origin and along the x-axis. Determine an expression for the total electric field at a point A (r - 2.1L,0) in terms of q and L. Calculate the magnitude of the x-axis. Determine an expression -LO) (L0) (r.0) -q +q Subnmit Answer Tries 0/7 Consider the situation as described in the diagram. For each of the following statements,...
I understand that to find the magnitude electric field you find the electric field of both...
I understand that to find the magnitude electric field you
find the electric field of both charges, but why are they added
together? Because there’s other examples I have encountered where
you subtract them instead
(10 pts) 8. A negative point charge q--5.0x10 C is on the +x-axis at x 0.400 m and a positive point charge 16010° C is on the negative xais atx-0200 m. a) What are the magnitude and direction of the resultant electric field at the...
An electric field can be created by a single charge or a distribution of charges. The...
An electric field can be created by a single charge or a distribution of charges. The electric field a distance from a point charge has magnitude E = k|q'|/r^2. The electric field points away from positive charges and toward negative charges. A distribution of charges creates an electric field that can be found by taking the vector sum of the fields created by individual point charges. Note that if a charge is placed in an electric field created by q',...
Calculate the magnitude and direction of the net electric field at the origin due to the...
Calculate the magnitude and direction of the net electric field at the origin due to the following three charges: 1. +3.8 microC @ (-.50m, 0) 2. -2.2 microC @ (-.20m, 0.15m) 3. -1.5 microC @ (-.05m, -0.12) Include vector diagram of situation.
Distribution of Charges Part A Calculate the magnitude of the electric field at the origin due to the following distribution of charges: -q at (x,y) - (a,a),-q at (a,-a), +q at (-a,-a) and +q at (-a,a), where q = 3.25 × 10-7 C . and a 4.40 cm2 Submit Answer Tries 0/6 Part B Set up an 8-point compass at the origin, where north points along the positive y-axis, such as that shown in the diagram to the right. What...
Distribution of Charges Calculate the resnitude of the electric field et the ongin due to the fulv 'ng distribution of charqes: q dt(x,y: d d -q dt d d ,-q et d djand +q dt d d where q = 3.5ข x 10 7 Cend d =。10 cm Submit Anzwer Trics /10 set up an ¥ pomt oor pass at the rigin, where north pu nts d ony he postive γ axis, such as t at sho n n he...
For starters, calculate the
magnitude and direction of the electric field due only to
charge q1 at this point.
Down
Up
Left
Right
Incompatible units. No conversion found between "N" and the
required units.
Tries 0/10
Previous Tries
Calculate the magnitude and direction of the electric field
due only to charge q2 at this
point.
Down
Up
Left
Right
Tries 0/10
Calculate the magnitude and direction of the electric field
due only to charge q3 at this
point.
Down
Up...
The diagram below shows a dipole centered at the total electric field at A, when q = 5.05×10-7 C and L-49.2 cm. origin and along the x-axis. Determine an expression for the total electric field at a point A (r - 2.1L,0) in terms of q and L. Calculate the magnitude of the x-axis. Determine an expression -LO) (L0) (r.0) -q +q Subnmit Answer Tries 0/7 Consider the situation as described in the diagram. For each of the following statements,...
I understand that to find the magnitude electric field you
find the electric field of both charges, but why are they added
together? Because there’s other examples I have encountered where
you subtract them instead
(10 pts) 8. A negative point charge q--5.0x10 C is on the +x-axis at x 0.400 m and a positive point charge 16010° C is on the negative xais atx-0200 m. a) What are the magnitude and direction of the resultant electric field at the...
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