1840 United States presidential election in Georgia

Election in Georgia

1840 United States presidential election in Georgia

← 1836 October 30 - December 2, 1840 1844 →
 
Nominee William Henry Harrison Martin Van Buren
Party Whig Democratic
Home state Ohio New York
Running mate John Tyler none
Electoral vote 11 0
Popular vote 40,339 31,983
Percentage 55.78% 44.22%

County Results

Harrison

  50-60%
  60-70%
  70–80%
  80–90%
  90–100%

Van Buren

  50-60%
  60-70%
  70-80%
  80-90%
  90-100%

Tie (50%)

  


President before election

Martin Van Buren
Democratic

Elected President

William Henry Harrison
Whig

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The 1840 United States presidential election in Georgia took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose 11 representatives, or electors to the Electoral College, who voted for President and Vice President.

Georgia voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Georgia by a margin of 11.56%.

Results

United States presidential election in Georgia, 1840[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Whig William Henry Harrison of Ohio John Tyler of Virginia 40,339 55.78% 11 100.00%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 31,983 44.22% 0 0.00%
Total 72,322 100.00% 11 100.00%

References

  1. ^ "1840 Presidential General Election Results - Georgia". U.S. Election Atlas. Retrieved December 23, 2013.


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